The following sample data represent the gasoline mileages (in miles per gallon) determined for cars in a particular weight class:
25.1 29.0 34.5 35.7 37.9 34.9 24.3 26.6 27.3 32.0 30.0 34.5 35.3 33.5 36.6 34.8 16.2 13.1 24.5 33.6 28.0 33.9 30.7 32.0 37.7 21.1 31.2 35.6 34.4 25.2 35.9 18.3 29.4 29.5 34.8 29.4 26.4 38.8 36.0 28.7 23.4 35.3 33.7 38.1 28.6 34.2 34.8 39.2 39.9 36.8
1. Using 10 mpg as the lower limit of the first class interval, construct a histogram with intervals of width 5 mpg. What is the midpoint of the cell with the largest number of observations (the mode).
2. Is the histogram skewed to the left or right or symmetric?
Answers
Answer:
1) Explanation and figure attached below.
2) For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.
Step-by-step explanation:
Part 1
If we order the data from the smallest to the largest we got:
13.1 , 16.2 ,18.3 , 21.1 , 23.4 , 24.3 , 24.5
25.1 , 25.2 , 26.4 , 26.6 ,27.3 ,28 , 28.6 ,
28.7 , 29 , 29.4 , 29.4 , 29.5 , 30 , 30.7
31.2 , 32 , 32 , 33.5 , 33.6 , 33.7 , 33.9
34.2 , 34.4 , 34.5 , 34.5 , 34.8 , 34.8
34.8 , 34.9 , 35.3 , 35.3 , 35.6 , 35.7
35.9 , 36 , 36.6 , 36.8 , 37.7 , 37.9
38.1 , 38.8 , 39.2 , 39.9
For this case if we use 10 mpg as the lower limit and with a width of 5 mpg for the intervals we have the following table:
Interval Frequency Midpoint
[10-15) 1 12.5
[15-20) 2 17.5
[20,25) 4 22.5
[25-30) 12 27.5
[30-35) 17 32.5
[35-40) 14 37.5
Total 50
So the histogram is on the figure attached. The midpoint for the class interval with the largest number of observations is 32.5.
Part 2
For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.
Final answer:
To construct the histogram, determine the class intervals and count the number of observations in each interval. The midpoint of the cell with the largest number of observations is approximately 37.45 mpg.
Explanation:
To construct a histogram with intervals of width 5 mpg, we start by determining the class intervals. Given that the lower limit of the first interval is 10 mpg, the class intervals can be calculated as follows: 10-14.9, 15-19.9, 20-24.9, 25-29.9, 30-34.9, 35-39.9. Next, we count the number of observations falling into each interval:
10-14.9: 3
15-19.9: 6
20-24.9: 5
25-29.9: 7
30-34.9: 9
35-39.9: 10
The cell with the largest number of observations is the one corresponding to the interval 35-39.9 mpg. To calculate the midpoint of this interval, we take the average of the lower and upper limits: (35 + 39.9) / 2 = 37.45. Therefore, the midpoint of the cell with the largest number of observations (the mode) is approximately 37.45 mpg.
Related Questions
Step1: find the Laplace transform of the solution Y(s).Y(s). Write the solution as a single fraction in s
Answers
Complete Question :
The question is shown on the first uploaded image
Answer:
The solution is the second uploaded image
Step-by-step explanation:
The step by step explanation is on the third, fourth and fifth uploaded image
Let D be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².
Write six different triple iterated integrals for the volume of D. Evaluate one of the integrals.
Answers
Answer:
∫∫∫1 dV=4\sqrt{3}π
Step-by-step explanation:
From Exercise we have
z=6-x^{2}-y^{2}
z=x^{2}+y^{2}
we get
2z=6
z=3
x^{2}+y^{2}=3
We use the polar coordinates, we get
x=r cosθ
y=r sinθ
x^{2}+y^{2}&=r^{2}
r^{2}=3
We get at the limits of the variables that well need for our integral
x^{2}+y^{2}≤z≤3
0≤r ≤\sqrt{3}
0≤θ≤2π
Therefore, we get a triple integral
\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA
=\int \int \left(z|_{x^2+y^2}^{3} \right) dA
=\int \int\ \left(3-(x^2+y^2) \right) dA
=\int \int\ \left(3-r^2 \right) dA
=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ
=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} 1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ
=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}} dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ
=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ
=3\sqrt{3} ·2π-\sqrt{3}·2π
=4\sqrt{3}π
We get
∫∫∫1 dV=4\sqrt{3}π
We find the volume of the region D bounded by the paraboloids z = 6 - x² - y² and z = x² + y² by setting up triple iterated integrals. Six different integrals are presented, and one is evaluated using cylindrical coordinates. The volume is determined to be 9π.
To find the volume of the region D bounded by the paraboloids z = 6 - x² - y² and z = x² + y², we need to set up triple iterated integrals.
The intersection of the two surfaces occurs when 6 - x² - y² = x² + y²,
which simplifies to 6 = 2(x² + y²) or x² + y² = 3, defining a circle of radius √3 in the xy-plane.
Possible Triple Iterated Integrals
Here are six different triple iterated integrals to find the volume of the region D:
[tex]\int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}} \int_{x^2+y^2}^{6-x^2-y^2} dz \, dy \, dx = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{-\sqrt{3}\cos\theta}^{\sqrt{3}\cos\theta} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3-y^2}}^{\sqrt{3-y^2}} \int_{y^2+x^2}^{6-y^2-x^2} dz \, dx \, dy \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{-\sqrt{3}\cos\theta}^{\sqrt{3}\cos\theta} \int_{x^2}^{6-x^2-\theta} dz \, dx \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{-\sqrt{3}}^{\sqrt{3}} \int_{y-x}^{y+x} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex]Evaluating One of the Integrals
Let's evaluate the triple iterated integral in cylindrical coordinates:[tex]\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{r^2}^{6 - r^2} r \, dz \, dr \, d\theta[/tex]
First, integrate with respect to z:[tex]\int_{r^2}^{6 - r^2}\, dz = \left[ z \right]_{z=r^2}^{z=6-r^2}[/tex]
[tex]= (6-r^2) - (r^2)[/tex]
[tex]= 6-2r^2[/tex]
Next, integrate with respect to r:[tex]int_{0}^{\sqrt{3}} r(6 - 2r^2) dr = \int_{0}^{\sqrt{3}} (6r - 2r^3) dr[/tex]
[tex]= \left[ 3r^2 - \frac{1}{2}r^4 \right]_{r=0}^{r= \sqrt{3}}[/tex]
[tex]= \left[ 3(3) - \frac{1}{2}(9) \right][/tex]
= 9 - 4.5
= 4.5
Finally, integrate with respect to θ:[tex]\int_{0}^{2\pi} 4.5 \, d\theta = 4.5 \cdot 2\pi = 9\pi[/tex]
So the volume of the region D is 9π.
What is the standard deviation (s) of the following set of scores? 10, 15, 12, 18, 19, 16, 12
Answers
Answer:
standard deviation =3.11
Step-by-step explanation:
data given is ungrouped data
standard deviation =√ [∑ (x-μ)² / n]
mean (μ)=∑x/n
= [tex]\frac{10+15+12+18+19+16+12}{7}[/tex]
=14.57
x-μ for data 10, 15, 12, 18, 19, 16, 12 will be
-4.57, 0.43, -2.57, 3.43, 4.43,1.43, -2.57
(x-μ)² will be 20.88, 0.1849, 6.6049, 11.7649,19.6249, 2.0449, 6.6049
∑ (x-μ)² will be = 67.7049
standard deviation = √(67.7049 / 7)
=3.11
Final answer:
The standard deviation for the given data is approximately 3.11.
Explanation:
To calculate the standard deviation (s) of the set of scores 10, 15, 12, 18, 19, 16, 12, you would follow these steps:
Calculate the mean score.Calculate the deviations from the mean for each score.Square each deviation from the mean.Calculate the mean of these squared deviations (this is the variance).Take the square root of the variance to find the standard deviation.Here's how to carry out each step with the provided scores:
The mean score is (10 + 15 + 12 + 18 + 19 + 16 + 12) / 7 = 102 / 7 ≈ 14.57.The deviations from the mean are -4.57, 0.43, -2.57, 3.43, 4.43, 1.43, -2.57.The squared deviations are 20.88, 0.18, 6.60, 11.76, 19.63, 2.04, 6.60.The variance is (20.88 + 0.18 + 6.60 + 11.76 + 19.63 + 2.04 + 6.60) / 7 ≈ 9.67.The standard deviation is the square root of 9.67, which is approximately 3.11.Therefore, the standard deviation of the scores is about 3.11.
Use the Pythagorean theorem to determine which of the following give the measures of the legs and hypotenuse of a right triangle. Which apply. 3,4,5. Or. 4,11,14. Or. 9,14,17. Or 8,14,16. Or. 8,15,17
Answers
Answer: 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
Step-by-step explanation:
In order for the measures of the legs and hypotenuse given to form a right angle triangle, they must be Pythagorean triples. A Pythagoras triple is a set of numbers that perfectly satisfy the Pythagorean theorem. The Pythagorean theorem is expressed as
Hypotenuse² = opposite side² + adjacent side². We will apply the theorem to each set of numbers given.
1) 3, 4, 5
5² = 3² + 4² = 9 + 16
25 = 25
It is a Pythagorean triple
2) 4, 11, 14
14² = 11² + 4² = 121 + 16
196 = 137
It is a Pythagorean triple
3) 9, 14, 17
17² = 14² + 9² = 196 + 81
289 = 277
It is not a Pythagorean triple
4) 8, 14, 16
16² = 14² + 8² = 196 + 64
256 = 260
It is not a Pythagorean triple
5) 8, 15 , 17
17² = 15² + 8²
289 = 225 + 64
289 = 289
It is a Pythagorean triple
Therefore, 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vector?
Answers
Answer:
[tex]\large\boxed {1.59 mm}[/tex]Explanation:
1. Given vector:
length: 4.00 mm = magnitude of the vectorangle: 23.5º north of east = 23.5º from the x-axys (counterclockwise)2. y-component
The y-component may be determined using the sine ratio, the angle from the x-axys (counterclockwise direction), and the magnitude of the vector.
sine (23.5º) = y-component / magnitudey-component = magnitude × sine (23.5º) = 4.00 mm × sine (23.5º) = 1.59 mm.[tex]\large\boxed{y-component = 1.59 mm}[/tex]Find an equation of the largest sphere that is centered at (5,4,9) and has interior contained in the first octant.
Answers
Answer:
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 16[/tex]
Step-by-step explanation:
The general equation of a sphere is as follows:
[tex](x - x_{c})^{2} + (y - y_{c})^{2} + (z - z_{c})^{2} = r^{2}[/tex]
In which the center is [tex](x_{c}, y_{c}, z_{c})[/tex], and r is the radius.
In this problem, we have that:
[tex]x_{c} = 5, y_{c} = 4, z_{c} = 1[/tex]
So
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = r^{2}[/tex]
Interior contained in the first octant:
The first octant is bounded by:
The xy plane, in which z is 0. The distance from the center of the sphere to the xy plane is 9.
The xz plane, in which y is 0. The distance from the center of the sphere to the xz plane is 4.
The yz plane, in which x is 0. The distance from the center of the sphere to the yz plane is 5.
This means that if the radius is higher than four, the sphere will cross into a different octant.
So the radius for the largest sphere is 4.
The equation is
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 4^{2}[/tex]
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 16[/tex]
If the interest rate is 7%, how many years will it take for your bank balance to double from $1,000 to $2,000?Enter the following data into your calculator:
Answers
It takes 10.3 years for your bank balance to double from $1,000 to $2,000.
Given that,
The interest rate is 7%.
Principal amount = $1000
Final amount = $2000
Used the formula for the time,
A = P (1 + r)ⁿ
Where, A = Final amount
P = Principal amount
r = interest rate
n = number of years
Substitute all the values,
2000 = 1000 (1 + 0.07)ⁿ
2000/1000 = (1.07)ⁿ
2 = (1.07)ⁿ
Take natural logs on both sides,
ln 2 = n ln (1.07)
0.69 = n × 0.067
n = 0.69/0.067
n = 10.3 years
Therefore, the time for your bank balance to double from $1,000 to $2,000 is 10.3 years.
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It will take approximately 10.1351 years for your bank balance to double from $1,000 to $2,000 at an interest rate of 7%.
To calculate the number of years it will take for your bank balance to double from $1,000 to $2,000 at an interest rate of 7%, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the final amount (in this case, $2,000),
P is the initial principal (in this case, $1,000),
r is the annual interest rate (7% or 0.07 as a decimal),
n is the number of times interest is compounded per year (we'll assume it's compounded annually),
and t is the number of years.
We can rearrange the formula to solve for t:
t = (log(A/P) / log(1 + r/n)) / n
Plugging in the values:
A = $2,000
P = $1,000
r = 0.07
n = 1 (since it's compounded annually)
t = (log(2,000/1,000) / log(1 + 0.07/1)) / 1
Simplifying the expression:
t = (log(2) / log(1.07)) / 1
Using a calculator to evaluate the logarithms:
t ≈ (0.3010 / 0.0296) / 1
t ≈ 10.1351 / 1
t ≈ 10.1351
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What are the rectangular coordinates of the point whose cylindrical coordinates are (r=9, θ=2π3, z=3)(r=9, θ=2π3, z=3) ?
Answers
Answer:
The point is [tex](-\frac{9}{2},\frac{9\sqrt{3}}{2},3)[/tex] in rectangular coordinates.
Step-by-step explanation:
To convert from cylindrical to rectangular coordinates we use the relations
[tex]x=r \cdot cos(\theta)\\y=r\cdot sin(\theta)\\z=z[/tex]
To convert the point [tex](9,\frac{2}{3}\pi ,3)[/tex] from cylindrical to rectangular coordinates we use the above relations
Since [tex]r=9[/tex], [tex]\theta=\frac{2}{3} \pi[/tex], and [tex]z=3[/tex],
[tex]x=r \cdot cos(\theta)=9\cdot cos(\frac{2}{3}\pi )=-\frac{9}{2}[/tex]
[tex]y=r\cdot sin(\theta)=9\cdot sin(\frac{2}{3} \pi )=\frac{9\sqrt{3}}{2}[/tex]
[tex]z=z=3[/tex]
Thus, the point is [tex](-\frac{9}{2},\frac{9\sqrt{3}}{2},3)[/tex] in rectangular coordinates.
One hundred students were given an Algebra test. A random sample of ten students was taken out of class of 800 enrolled students. The time it took each student to complete the test is recorded below. 22.2, 23.7, 16.8, 18.3, 19.7, 16.9, 17.2, 18.5, 21.0, and 19.7
a. Find the mean, variance and standard deviation for this sample of ten students
b. Construct a 95% confidence interval for the population mean time to complete this Algebra test.
c. Test if the population mean time to complete the test is 22.5 minutes.
Answers
Answer:
a) [tex]\bar X= \frac{\sum_{i=1}^n X_i}{10} =19.4[/tex]
[tex] s^2 = \frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}=5.438[/tex]
[tex] s= \sqrt{5.438}=2.332[/tex]
b) [tex]19.4-2.262\frac{2.332}{\sqrt{10}}=17.732[/tex]
[tex]19.4+2.262\frac{2.332}{\sqrt{10}}=21.068[/tex]
So on this case the 95% confidence interval would be given by (17.732;21.068)
c) [tex]t=\frac{19.4-22.5}{\frac{2.332}{\sqrt{10}}}=-4.203[/tex]
Since is a two-sided test the p value would be:
[tex]p_v =2*P(t_{9}<-4.203)=0.00230[/tex]
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 22.5 minutes.
Step-by-step explanation:
Part a
We have the following data given: 22.2, 23.7, 16.8, 18.3, 19.7, 16.9, 17.2, 18.5, 21.0, and 19.7
We can calculate the sample mean with this formula:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{10} =19.4[/tex]
And the sample variance with this formula:
[tex] s^2 = \frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}=5.438[/tex]
And the sample deviation would be just the square root of the sample variance
[tex] s= \sqrt{5.438}=2.332[/tex]
Part b
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X =19.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=2.332 represent the sample standard deviation
n=10 represent the sample size
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]
Now we have everything in order to replace into formula (1):
[tex]19.4-2.262\frac{2.332}{\sqrt{10}}=17.732[/tex]
[tex]19.4+2.262\frac{2.332}{\sqrt{10}}=21.068[/tex]
So on this case the 95% confidence interval would be given by (17.732;21.068)
Part c
Null hypothesis:[tex]\mu =22.5[/tex]
Alternative hypothesis:[tex]\mu \neq 22.5[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{19.4-22.5}{\frac{2.332}{\sqrt{10}}}=-4.203[/tex]
P-value
Since is a two-sided test the p value would be:
[tex]p_v =2*P(t_{9}<-4.203)=0.00230[/tex]
Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 22.5 minutes.
Final answer:
The mean, variance, and standard deviation of a sample of ten students' test completion times are 19.0 minutes, 3.3 minutes² , and 1.82 minutes, respectively.
Explanation:
a. To find the mean, add up all the times and divide by the number of students: (22.2 + 23.7 + 16.8 + 18.3 + 19.7 + 16.9 + 17.2 + 18.5 + 21.0 + 19.7) / 10 = 19.0 minutes.
To find the variance, subtract the mean from each time, square the differences, and find the average: [(22.2-19.0)² + (23.7-19.0)² + ... + (19.7-19.0)²] / 10 = 3.3 minutes².
To find the standard deviation, take the square root of the variance: √(3.3) = 1.82 minutes.
b. To construct a 95% confidence interval, we need to know the critical value for a sample size of 10. The critical value for a 95% confidence interval with 10 degrees of freedom is 2.262.
The margin of error can be calculated as the critical value multiplied by the standard deviation: 2.262 * (1.82 / √10) = 1.29 minutes. The confidence interval is then [19.0 - 1.29, 19.0 + 1.29] = [17.71, 20.29].
c. To test if the population mean time to complete the test is 22.5 minutes, we can use a t-test. The t-value can be calculated as the difference between the sample mean and the hypothesized population mean divided by the standard deviation divided by the square root of the sample size: (19.0 - 22.5) / (1.82 / √10) = -5.44. T
he degrees of freedom for this test is 9. Using a t-distribution table, we find that the critical t-value for a two-tailed test at a significance level of 0.05 is approximately ±2.262. Since -5.44 is outside of this range, we can reject the null hypothesis that the population mean time is 22.5 minutes.
In a survey of 447 registered voters, 157 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 27% of registered voters wish to see her defeated. Does the 98% confidence interval for the proportion support this claim? (Hint: you should first construct the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated.) (0.299, 0.404)
A. No
B. Yes
Answers
Let p be the true proportion of registered voters wish to see Mayor Waffleskate defeated.
As per given , we have
[tex]H_0: p\leq0.27\\\\ H_a: p >0.27[/tex]
Sample size : n= 447
Number of of registered voters wish to see Mayor Waffleskate defeated = 157
I.e. sample proportion : [tex]\hat{p}=\dfrac{157}{447}\approx0.3512[/tex]
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = sample proportion
z* = critical z-value.
Critical z-value for 98% confidence interval is 2.33. (By z-table)
Then, the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated will be :
[tex]0.3512\pm2.33\sqrt{\dfrac{0.3512(1-0.3512)}{447}}\\\\=0.3512\pm (2.33)(0.022577656)\\\\=0.3512\pm 0.05260593848\\\\=(0.3512-0.05260593848,\ 0.3512+0.05260593848)\\\\=(0.29859406152,\ 0.40380593848)\approx(0.299,\ 0.404)[/tex]
Since the 0.27 < 0.299 , it means 0.27 does not belong to the above confidence interval.
So , we reject the null hypothesis ([tex]H_0[/tex]).
So , 98% confidence interval does not support the claim.
Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine randomly selected individuals are given below:
108.6 117.4 128.4 120.0 103.7 112.0 98.3 121.5 123.2
(a) What is the median of the reported blood pressure values?(b) Suppose the blood pressure of the second individual is 117.9 rather than 117.4 (a small change in a single value). What is the new median of the reported values?(c) What does this say about the sensitivity of the median to rounding or grouping in the data?A. When there is rounding or grouping, the median can be highly sensitive to small change.B. When there is rounding or grouping, the median is only sensitive to large changes.C. When there is rounding or grouping, the median is not sensitive to small changes.
Answers
Answer:
a) 117.4
b) 117.9
c) Option A) When there is rounding or grouping, the median can be highly sensitive to small change
Step-by-step explanation:
We are given the following data set in the question:
108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
n = 9
a) Median of the reported blood pressure values
Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4
Median =
[tex]\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4[/tex]
b) New median of the reported values
Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2
Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4
New Median =
[tex]\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9[/tex]
c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.
Option A) When there is rounding or grouping, the median can be highly sensitive to small change
The median of the reported blood pressure values is 117.4. If the blood pressure of the second individual is changed to 117.9, the new median would be 117.9. This shows that the median is not sensitive to small changes in rounding or grouping.
Explanation:(a) What is the median of the reported blood pressure values?
To find the median, we need to arrange the blood pressure values in numerical order: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4. Since there are 9 values, the middle value is the 5th value, which is 117.4.
(b) Suppose the blood pressure of the second individual is 117.9 rather than 117.4. What is the new median of the reported values?
If we change the second individual's blood pressure to 117.9, the new order is: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4. Now, there are 9 values and the middle value is the 5th value, which is 117.9. So the new median would be 117.9.
(c) What does this say about the sensitivity of the median to rounding or grouping in the data?
C. When there is rounding or grouping, the median is not sensitive to small changes.
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Harry Potter approaches with a strange bag full of balls, numbered 1 to k. As you reach in to pick one, he notes that they are not all equally likely because of magic: ball 1 is least likely to be chosen, with probability c, where c is some constant. Ball 2 has probability 2c, Ball 3 has probability 3c, . . . , Ball k − 1 has probability (k − 1)c, and Ball k has probability kc.
1. What is the expected value of the ball number you pick? Your answer can’t use the constant c, but will use k.
Answers
Answer:
[k*(k+1)*(2*k+1)] / 6
Step-by-step explanation:
We have balls numbered as: 1, 2, 3, ... , k with probabilities as: c, 2*c, 3*c, ... , k*c
Let Y be the discrete random variable defined as: Y = ball number
We know that Expected value of discrete Random Variable is:
E[X] = Σ₁ⁿ xₐ*f(xₐ) ,where f(xₐ) is probability of xₐ
then,
E[Y] = 1*c + 2*2*c + 3*3*c + ... + k*k*c
E[Y] = c*(1 + 2*2 + 3*3 + ... + k*k)
E[Y] = c*(1^2 + 2^2 + 3^2 + ... + k^2)
consider c = 1 (because it's constant so you can suppose any you wish)
E[Y] = 1^2 + 2^2 + 3^2 + ... + k^2
using formula of first n squares natural numbers (as attached picture)
E[Y] = [k*(k+1)*(2*k+1)] / 6
The teacher recorded the mean and median of the hourly wage for each student. Unfortunately, he forgot to label them. The numbers he wrote down were: $11.25/hour and $9.38/hour. Which would be the mean and which would be the median
Answers
Step-by-step explanation:
We can not exactly predict the values of mean and median of the data un till and unless we know about the skewness of the data.
Skewness represents the asymmetry or tapering in the distribution of data sample. If skewness is
Negative skew: median > mean:
Positive skew: mean > median :
Although this generalization is not always true.
State all possible names for each figure.
Answers
Answer:
Step-by-step explanation:
square
quadrilateral - all of these
polygon - squares
trapezoid - others - non square
rhombus - squares
hope this helps.
A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain to a height of 4 m?
Answers
Answer:
[tex] W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J[/tex]
Step-by-step explanation:
Data Given: m = 70 kg , g = 9.8 ms^-2, h =10m.
For this case we can use the following formula:
[tex] W = \int_{x_i}^{x_f} F(x) dx[/tex]
For this case we need to find an expression for the force in terms of the distance. And since on this case the total distance is 10 m long we can write the expression like this:
[tex] F(x) = \frac{ma}{10m}= \frac{mg}{10m} x[/tex]
The only acceleration on this case is the gravity and if we replace the values given we got:
[tex] \frac{70 kg *9.8 m/s^2}{10m} x=68.6 x\frac{kg}{s^2}[/tex]
Now we can find the required work with the following integral:
[tex] W= 68.6 \frac{kg}{s^2} \int_{0}^4 x dx[/tex]
[tex] W= 34.3 \frac{kg}{s^2} x^2 \Big|_0^4[/tex]
[tex] W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J[/tex]
The amount of work that is required to raise one end of the chain is 548.8 Joules.
Given the following data:
Length of chain = 10 meters.Mass of chain = 70 kg.Height = 4 meters.To calculate the amount of work that is required to raise one end of the chain:
How to calculate the work done.We would solve for the magnitude of the force acting on the chain with respect to the distance and this is given by this expression:
[tex]Force = \frac{mgx}{10} \\\\Force = \frac{70 \times 9.8 \times x}{10}[/tex]
Force = 68.6x Newton.
Now, we can calculate the amount of work by using this formula:
[tex]W=\int\limits^{x_2}_{x_1} F({x}) \, dx \\\\W= 68.6 \int\limits^{4}_{0} x \, dx\\\\W= 34.3 x^2 |^4_0\\\\W=34.3 [4^2 -0^]\\\\W=34.3 \times 16[/tex]
W = 548.8 Joules.
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The upper arm length of females over 20 years old in a country is approximately Normal with mean 35.8 centimeters (cm) and standard deviation 2.5 cm. Use the 68-95-99.7 rule to answer the following questions. (Enter your answers to one decimal place.) (a) What range of lengths covers almost all (99.7%) of this distribution? From 33.3 Incorrect: Your answer is incorrect. cm to 38.3 Incorrect: Your answer is incorrect. cm (b) What percent of women over 20 have upper arm lengths less than 33.3 cm? 2.5 Incorrect: Your answer is incorrect. %
Answers
Answer:
a) The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.
b) 16% of women over 20 have upper arm lengths less than 33.3 cm.
Step-by-step explanation:
The Empirical Rule(68-95-99.7 Rule) states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 35.8 cm
Standard deviation = 2.5 cm
(a) What range of lengths covers almost all (99.7%) of this distribution?
This range is from 3 standard deviations below the mean to three standard deviations above the mean.
So from 35.8 - 3*2.5 = 28.3 cm to 35.8 + 3*2.5 = 43.3 cm
The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.
(b) What percent of women over 20 have upper arm lengths less than 33.3 cm?
68% of the women over 20 have upper arm length between 33.3 cm and 38.3 cm. The other 32% have upper arm length lower than 33.3 cm or higher than 38.3. The distribution is symmetric, so 16% of the have upper arm length lower than 33.3 cm and 16% have upper arm length higher than 38.3 cm
So 16% of women over 20 have upper arm lengths less than 33.3 cm.
The powerful survival impulse that leads infants to seek closeness to their caregivers is called:A)attachment.B)imprinting.C)habituation.D)assimilation.E)the rooting reflex
Answers
Answer:
A. Attachment
Step-by-step explanation:
The powerful survival impulse that leads infants to seek closeness to their caregivers is called Attachment. The infant can count on the caregiver possibly parent(s) for care which gives the infant a solid foundation for dependence and survival.
Final answer:
The instinctual behavior that drives infants to seek closeness with their caregivers is known as A) attachment. It's fostered by reflexes that ensure physical contact and is crucial for an infant's survival, ensuring they receive the necessary care, protection, and opportunity to develop securely.
Explanation:
The powerful survival impulse that leads infants to seek closeness to their caregivers is called A) attachment. This is an intrinsic part of human development and is crucial for the infant's survival. Infants have a set of innate behaviors and reflexes that promote closeness and contact with their caregivers, such as the Moro reflex and the grasping reflex, which help the infant to hold onto the caregiver and thus reduce the risk of falling.
Additionally, behaviors such as crying and the sucking reflex are instinctive methods for infants to express needs and receive care. Furthermore, the rooting reflex is an instinctive behavior that helps the infant find the nipple to feed by touching. John Bowlby's evolutionary theory underscores the importance of attachment by suggesting that the ability to maintain proximity to an attachment figure would have increased the chances of an infant surviving to reproductive age.
Attachments are not just reactions to the provision of food and warmth by the caregivers but are biological imperatives that ensure an infant remains close to those who provide security, learning, and protection, thereby enhancing their chance of survival.
A study group is to be selected from 5 freshmen, 7 sophomores, and 4 juniors. a) If a study group is to consist of 2 freshmen, 3 sophomores, and 1 junior, how many different ways can the study group be selected? b) If a study group consisting of 6 students is selected, what is the probability that the group will consist of 2 freshmen, 3 sophomores, and 1 junior?
Answers
Answer:
175/1001
Step-by-step explanation:
(a.)
Freshman: 5 Combination 2
sophomores: 7 Combination 3
Junior: 4 Combination 1
The combination of three groups :
=5c2 × 7c3 × 4c1
=10 × 35 × 4
=1400 ways
(b.)
=5+7+4=16
=2+3+1
Which is (16 combination 6)
The probability will be:
=(5c2 × 7c3 × 4c1) / 16c6
=1400/8008
=175/1001
The groups can be formed in multiple ways, calculated by the combination formula, and the probability of forming a specific group is calculated by dividing the number of ways to form that specific group by the total number of possible groups.
Explanation:The question is about how many unique ways a study group can be formed from a given set of students and determining the probability of a specific group formation.
To solve Part (a), we use the concept of combinations. The number of ways to select 2 freshmen out of 5 is 5C2, for 3 sophomores out of 7 is 7C3, and for 1 junior out of 4 is 4C1. Multiply these together to get the total number of ways, which is (5C2)*(7C3)*(4C1).
For Part (b), the total number of ways to select a group of 6 students from 16 is 16C6. The probability of selecting 2 freshmen, 3 sophomores, and 1 junior, is the result from Part (a) divided by this total number, or [(5C2)*(7C3)*(4C1)] /(16C6).
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A statistician controls ____________ by establishing the risk he or she is willing to take in terms of rejecting a true null hypothesis.
a) Alpha
b) beta
c) mean
d) standard deviation
Answers
Answer:
a) Alpha
Step-by-step explanation:
The correct option is alpha because alpha known as type I error is the probability of reject the null hypothesis when null hypothesis is true. If we take alpha 5%, it means that we are taking 5 out of 100 chance of rejecting the null hypothesis when it is true.
So, statistician controls alpha by establishing the risk of rejecting a null hypothesis when its true.
what is the answer for 6+3×2
Answers
Pemdas
3*2=6
6+6=12
6 + 6 = 12
Answer = 12
You randomly select k integers between 1 and 100, inclusive. What is the smallest k that guarantees that at least one pair of the selected integers will sum to 101? Prove your answer.
Answers
Answer:
51
Step-by-step explanation:
The possible components to sum up to 101 can only be divided into 2 groups, 1 is larger than 50 and the other is less than or equals to 50
For example
50 + 51 = 101
52 + 49 = 101
60 + 41 = 101
...
99 + 2 = 101
100 + 1 = 101
Therefore, the worst case scenario is to pick all numbers from only 1 group, either all number less than or equal to 50, which there are 50 of them from 1 to 50, or greater than 50, which there are 50 of them from 51 to 100.
So k has to be at least 51 to guarantee that at least one pair of the selected integers will sum to 101
The smallest value of k that guarantees at least one pair of the selected integers will sum to 101 is 100.
Explanation:In order to guarantee that at least one pair of the selected integers will sum to 101, we need to find the largest value of k under which it is still possible for all pairs of integers to have a sum less than or equal to 100.
Let's consider the scenario where k is equal to 99. We can select 99 integers between 1 and 100, and it is still possible for none of the pairs to have a sum of 101. Every integer can be paired with one other integer to form a sum less than or equal to 100 (e.g., 1 and 100, 2 and 99, 3 and 98, and so on).
However, if we select one more integer to make k equal to 100, there will be at least one pair that adds up to 101, since we have one extra integer that cannot form a sum less than 101 with any of the other integers.
Therefore, the smallest value of k that guarantees at least one pair of the selected integers will sum to 101 is 100.
Luis spent $745.10 on 13 new file cabinet for his office. Small file cabinets cost $43.50 and large file cabinets cost $65.95. Write and solve a system of equations to find the number of smal cabinets and large cabinets he purchased.
Answers
Answer: he purchased 5 small file cabinets and 8 large file cabinets.
Step-by-step explanation:
Let x represent the number of small file cabinets that he purchased.
Let y represent the number of large file cabinets that he purchased.
Luis bought 13 new small and large file cabinets. This means that
x + y = 13
Small file cabinets cost $43.50 and large file cabinets cost $65.95. He spent a total of $745.10. This means that
43.5x + 65.95y = 745.1 - - - - - - - -1
Substituting x = 13 - y into equation 1, it becomes
43.5(13 - y) + 65.95y = 745.1
565.5 - 43.5y + 65.95y = 745.1
- 43.5y + 65.95y = 745.1 - 565.5
22.45y = 179.6
y = 179.6/22.45
y = 8
x = 13 - y = 13 - 8
x = 5
The problem can be solved by creating a system of linear equations based on the given information, then solved using methods such as substitution or elimination to find out the number of small and large cabinets.
Explanation:This is a problem related to the system of linear equations. So let's define the variables: Let 's' represent the number of small cabinets and 'l' the number of large cabinets. So we form two equations, one based on the total amount spent, and the other based on the total number of cabinets.
The first equation is $43.50s + $65.95l = $745.10, presenting the total amount of money spent by Luis. The second one is s + l = 13, representing the total number of cabinets.
Now, we can solve these equations simultaneously using any method, substitution or elimination, to find the values of 's' and 'l'.
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The acceleration due to gravity on Earth is 32 ft/sec2. A tomato is dropped from 100 feet above the ground. Give units in your answers. (a) At what speed does the tomato hit the ground? (b) How long does it take for the tomato to travel the last 10 feet? Give your answer as a decimal approximation with units.
Answers
Final answer:
The tomato hits the ground at a speed of approximately 80 ft/s. It takes approximately 1.58 seconds for the tomato to travel the last 10 feet.
Explanation:
To find the speed at which the tomato hits the ground, we can use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (which we want to find)u is the initial velocity (zero in this case)a is the acceleration due to gravity (-32 ft/s^2)s is the distance traveled (100 ft in this case)Plugging in these values, we get:
v^2 = 0 + 2(-32)(100)
v^2 = -6400
Taking the square root of both sides, we find that the speed at which the tomato hits the ground is approximately 80 ft/s.
To find the time it takes for the tomato to travel the last 10 feet, we can use the equation:
s = ut + (1/2)at^2
Where:
s is the distance traveled (10 ft in this case)u is the initial velocity (zero, as the tomato starts from rest)a is the acceleration due to gravity (-32 ft/s^2)t is the time it takes to travel the distancePlugging in these values, we get:
10 = 0 + (1/2)(-32)t^2
20 = -16t^2
t^2 = -20/16
Taking the square root of both sides, we find that the time it takes for the tomato to travel the last 10 feet is approximately 1.58 seconds.
Which operation do you think of when you see the word of
Answers
Answer:
division for me im probably wrong tbh
Step-by-step explanation:
Three data entry specialists enter requisitions into a computer. Specialist 1 processes 30 percent of the requisitions, specialist 2 processes 45 percent, and specialist 3 processes 25 percent. The proportions of incorrectly entered requisitions by data entry specialists 1, 2, and 3 are 0.03, 0.05, and 0.02, respectively. Suppose that a random requisition is found to have been incorrectly entered. What is the probability that it was processed by data entry specialist 1
Answers
Answer: =0.2465
Step-by-step explanation:
Firstly the probability that a random requisition is entered by data entry specialist 1 equals 0.3, by data
entry specialist 2 equals 0.45, by data entry specialist 3 equals 0.25.
probability(p1) =
(0.3 × 0.03
)/(0.3 × 0.03 + 0.45 × 0.05 + 0.25 × 0.02 )= 0.2465
A sample of 15 from a normal population yields a sample mean of 43 and a sample standard deviation of 4.7. What is the P—value that should be used to test the claim that the population mean is less than 45? a. 0.0608 b. 0.1216 c. 0.4696 d. 0.9392 e. The P—value cannot be determined from the given information.
Answers
Answer:
b. 0.1216
Step-by-step explanation:
Given that a sample of 15 from a normal population yields a sample mean of 43 and a sample standard deviation of 4.7.
We have to check the p value for the claim that mean <45
[tex]H_0: \mu =45\\H_a: \mu <45[/tex]
(Left tailed test for population mean)
Sample size n = 15
Sample mean = 45
Sample std dev s = 4.7
Since sample std deviation is being used, we use t test only
Std error of mean = [tex]\frac{s}{\sqrt{n} } \\=1.214[/tex]
Mean difference = 43-45 = -2
t statistic = mean difference/std error
= -1.176
df = n-1 = 14
p value = 0.1216
Write the equation in vertex form for the parabola with focus (0,5) and directrix y=
–
5.
Simplify any fractions.
Answers
Answer: [tex]x^{2} = 20y[/tex]
Step-by-step explanation:
The directrix given is vertical , so we will use the formula :
[tex](x-h)^{2}=4p(y-k)[/tex]
P is the distance between the focus , that is 5 - 0 = 5
Therefore : p = 5
(h,k) is the mid point between the focus and the directrix , that is
(h,k) = [tex](\frac{x_{1}+x_{2} }{2},\frac{y_{2}+y_{1}}{2})[/tex] = [tex](\frac{0+0}{2} , \frac{5-5}{2})[/tex] = [tex](0,0)[/tex]
Therefore:
h =0
k = 0
substituting into the formula : we have
[tex](x-h)^{2}=4p(y-k)[/tex]
[tex](x-0)^{2}[/tex] = 4(5)([tex]y-0)[/tex]
[tex]x^{2} = 20y[/tex]
Therefore : the equation in vertex form is [tex]x^{2} = 20y[/tex]
The position of a particle moving to the right on the x-axis is given by x(t), where x(t) is measured in inches and t is measured in minutes for 0≤t≤100. If y=x(t) is a linear function, which of the following would most likely give the best estimate of the speed of the particle, in inches per minute, at time t=20 minutes?
A. x(20)x(20)
B. x(20)20x(20)20
C. x(21)−x(19)x(21)−x(19)
The slope of the graph of y=x(t)
Answers
Answer:
The slope of the graph of y=x(t)
Step-by-step explanation:
The speed's equation is the derivative of the position.
Position
y = x(t)
Linear function.
So
y = ax(t)
In which a is the slope
The derivative of y is:
y = a
Which is the slope
This means that the speed of the particle is constant.
So the answer is:
The slope of the graph of y=x(t)
Find all values of m so that the function
y = emx
is a solution of the given differential equation. (Enter your answers as a comma-separated list.)
y' + 7y = 0
Answers
Answer:
[tex]m = -7[/tex]
Step-by-step explanation:
The objective is to find all values [tex]m[/tex] so that the function [tex]y=e^{mx}[/tex] is a solution of the differential equation [tex]y'+7y =0[/tex].
If [tex]y=e^{mx}[/tex] is a solution of the given differential equation, then it and its first derivative must satisfy the given equation. Let's calculate the derivative.
[tex]y = e^{mx} \implies y' = e^{mx} \overset{\text{Chain Rule}}{\cdot} (mx)' = me^{mx}[/tex]
Substituting [tex]e^{mx}[/tex] for [tex]y[/tex] and [tex]me^{mx}[/tex] for [tex]y'[/tex] in the equation gives
[tex]me^{mx} + 7e^{mx} = 0 \iff e^{mx}(m+7) = 0[/tex]
We can divide both sides by [tex]e^{mx}[/tex], since [tex]e^{mx} > 0, \; \forall x,m \in \mathbb{R} \implies e^{mx} \neq 0[/tex]. Thus,
[tex]m +7 = 0 \implies m = -7[/tex]
Therefore, the function [tex]y = e^{-7x}[/tex] is a solution of the differential equation [tex]y'+7y = 0.[/tex]
Exercise 3.23 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes. (a) What is the probability that their first child will have green eyes and the second will not?
Answers
Answer:
There is a 10.9375% probability that their first child will have green eyes and the second will not.
Step-by-step explanation:
We have these following probabilities:
0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.
(a) What is the probability that their first child will have green eyes and the second will not?
There is a 0.125 probability a child will have green eyes and an 1-0.125 = 0.875 probability a child will not have green eyes.
So
0.125*0.875 = 0.109375
There is a 10.9375% probability that their first child will have green eyes and the second will not.