"The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients, oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats and rice it should include in each box of cereal to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes 8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes 6 milligrams of A and 2 milligrams of B. An ounce of oats costs $0.05, and an ounce of rice costs $0.03. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis."

Answer:

[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]

-The sample is too small to make judgments about skewness or symmetry.

H0:[tex]\mu_{1}=\mu_{2}[/tex]

H1:[tex]\mu_{1} \neq \mu_{2}[/tex]

[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]

[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]

So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003      d2=1.337-1.322=0.015

d3=1.079-1.073=0.006     d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002   d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005      d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]

And for the sample deviation we can use the following formula:

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

[tex]\bar X_{1}=1.173[/tex] represent the mean for the operator 1

[tex]\bar X_{2}=1.174[/tex] represent the mean for the operator 2

[tex]s_{1}=0.1506[/tex] represent the sample standard deviation for the operator 1

[tex]s_{2}=0.1495[/tex] represent the sample standard deviation for the operator 2

[tex]n_{1}=8[/tex] sample size for the operator 1

[tex]n_{2}=8[/tex] sample size for the operator 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0:[tex]\mu_{1}=\mu_{2}[/tex]

H1:[tex]\mu_{1} \neq \mu_{2}[/tex]

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) like this:

[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]

Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{1}+n_{2}-2=8+8-2=14[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]

So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Answer:

[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]

Step-by-step explanation:

Circle and Square

We have a geometric construction as shown in the image below. We can see that

[tex]\displaystyle r+h=L[/tex]

Or, equivalently

[tex]\displaystyle h=L-r[/tex]

The triangle formed by r,h and L/2 is right, because the opposite side of the square is tangent to the circle at its midpoint. This means we can use Pythagoras's theorem:

[tex]\displaystyle r^2=h^2+\left(\frac{L}{2}\right)^2[/tex]

Replacing h

[tex]\displaystyle r^2=(L-r)^2+\left(\frac{L}{2}\right)^2[/tex]

Expanding squares

[tex]\displaystyle r^2=L^2-2Lr+r^2+\frac{L^2}{4}[/tex]

Simplifying

[tex]\displaystyle 2Lr=L^2+\frac{L^2}{4}[/tex]

Multiplying by 4

[tex]\displaystyle 8r=4L+L[/tex]

Joining terms

[tex]\displaystyle 8r=5L[/tex]

Solving for the ratio L/R as required

[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]

The  ratio of the side and the radius is 8:5

Since

r + h = L

So we can write h = L - r

Now we applied the Pythagoras theorem

[tex]r^2 = h^2 +( \frac{L}{2}) ^2\\\\r^2 = (L-r)^2 + ( \frac{L^}{2}) ^2\\\\r^2 = L^2- 2Lr+ r^2 + \frac{L^2}{4}\\\\ 2Lr = L^2 \frac{L^2}{4}\\\\[/tex]

Now

8r = 4L + L

8r = 5L

Learn more about the ratio here: https://brainly.com/question/13176103?referrer=searchResults

Answer:

There is a 99.44% probability that the squad will have at most 2 calls in an hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

In this problem, we have that:

2.83 rescues every eight hours.

What is the probability that the squad will have at most 2 calls in an hour?

This is

[tex]P = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

We have 2.83 rescues every 8 hours. So for an hour, we have [tex]\mu = \frac{2.83}{8} = 0.354[/tex]

So

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019[/tex]

[tex]P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485[/tex]

[tex]P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440[/tex]

So

[tex]P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944[/tex]

There is a 99.44% probability that the squad will have at most 2 calls in an hour.

Final answer:

The probability that the rescue squad will receive at most 2 calls in an hour is 0.9951

when the average is 2.83 calls every eight hours.

Explanation:

The number of rescue calls that a rescue squad receives is given to be a Poisson distribution with an average of 2.83 rescues every eight hours. To calculate the probability of receiving at most 2 calls in one hour, we first find the average number of rescues per hour by dividing the given average by eight, since there are eight hours in the period mentioned. This gives us an average rate ([tex]λ[/tex]) of 2.83 rescues / 8 hours = 0.35375 rescues per hour. The Poisson probability formula is:

[tex]P(X=k) = (e^{-λ} * λ^k) / k![/tex]

To find the probability of at most 2 calls, we sum the probabilities of 0, 1, and 2 calls:

[tex]P(X=0) = e^{-0.35375} * (0.353750)^0 / 0! = 0.7026 \\\\P(X=1) = e^{-0.35375} * (0.353751)^1 / 1! = 0.2485 \\\\P(X=2) = e^{-0.35375} * (0.353752)^2 / 2! = 0.0440[/tex]

Thus, the desired probability is the sum of these probabilities:

P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) ≈ 0.7026 + 0.2485 + 0.0440 ≈ 0.9951

After rounding to four decimal places we have that the probability that the squad will have at most 2 calls in an hour is 0.9951.

Answer:

[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]  

[tex]p_v =2*P(z>1.897)=0.0578[/tex]  

If we compare the p value obtained 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.  

Step-by-step explanation:

1) Data given and notation

n=250 represent the random sample taken

X=140 represent the number of heads obtained

[tex]\hat p=\frac{140}{250}=0.56[/tex] estimated proportion of heads

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that that one-Euro coins are biased, so the correct system of hypothesis are:  

Null hypothesis:[tex]p=0.5[/tex]  

Alternative hypothesis:[tex]p \neq 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

[tex]np_o =250*0.5=125>10[/tex]

[tex]n(1-p_o)=250*(1-0.5)=125>10[/tex]

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>1.897)=0.0578[/tex]  

If we compare the p value obtained 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.  

Answer

The answer and procedures of the exercise are attached in the following archives.

The information complete of the exercise is attached in the sheet.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

356 dies were examined by an inspection probe and 183 of these passed the probe. This neabs that [tex]n = 356, \pi = \frac{183}{356} = 0.514[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 - 1.96\sqrt{\frac{0.514*0.486}{356}} = 0.462[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 + 1.96\sqrt{\frac{0.514*0.486}{356}}{119}} = 0.566[/tex]

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).

Answer:Sin of ∠D = 39/89

Step-by-step explanation:

The diagram of the right angle DEF is shown in the attached photo.

To determine the ratio representing the sine of ∠D, we would apply the sine trigonometric ratio. It is expressed as

Sin# = opposite side/hypotenuse

Looking at the triangle

Opposite side = 39

Hypotenuse = 89

# = ∠D

Therefore

Sin of ∠D = 39/89

The sine of ∠D in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. So, in ΔDEF, sine of ∠D is FE/ED = 39/89.

In trigonometry, the sine of an angle in a right triangle is defined as the ratio of the length of the side that is opposite that angle, to the length of the hypotenuse. Considering this, the ratio representing the sine of ∠D in ΔDEF would be the length of side FE (opposite ∠D) divided by the length of side ED (the hypotenuse in ΔDEF because ∠F is a right angle and ED lies opposite to it).

Therefore, the ratio representing the sine of ∠D is FE/ED = 39/89.

Learn more about Trigonometry Ratio here:

https://brainly.com/question/32404697

#SPJ3

Answer: 482

Step-by-step explanation:

Formula to find the sample size is given by :-

[tex]n= (\dfrac{z^*\times \sigma}{E})^2[/tex]                                    (1)

, where z* = critical z-value (two tailed).  

[tex]\sigma[/tex] = Population standard deviation and E = Margin of error.

As per given , we have

Margin of error : E= 3

[tex]\sigma=40[/tex]

Confidence level = 90%

Significance level =[tex]\alpha=1-0.90=0.10[/tex]

Using z-table , the critical value for 90% confidence=[tex]z^*=z_{\alpha/2}=z_{0.05}=1.645[/tex]

Required minimum sample size = [tex]n= (\dfrac{(1.645)\times (40)}{3})^2[/tex]   [Substitute the values in formula (1)]

[tex]n=(21.9333333333)^2[/tex]

[tex]n=481.07111111\approx482[/tex]  [ Round to the next integer]

Hence, the number of observations required is closest to 482.

Answer:

0.3 or 30%

Step-by-step explanation:

Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :

[tex]P = 0.5*0.6 = 0.3[/tex]

The probability is 0.3 or 30%.

Answer:a) 5 sides

b) 9 sides

Step-by-step explanation:

a) Since the polygon is a regular polygon, all the interior angles are equal. If one exterior angle is 72 degrees, the interior angle would be

180 - 72 = 108 degrees(sum of the angles on a straight line is 180 degrees)

The sum of the interior angles of a polygon is expressed as

180(n - 2)

Where n represents the number of sides that the polygon has. Since the interior angles are equal, then

108n = 180(n - 2)

108n = 180n - 360

180n - 108n = 360

72n = 360

n = 360/72 = 5

b) If one exterior angle is 40 degrees, the interior angle would be

180 - 40 = 140 degrees

Since the interior angles are equal, then

140n = 180(n - 2)

140n = 180n - 360

180n - 140n = 360

40n = 360

n = 360/40 = 9

Answer:The value of the bulldozer after 3 years is $121950

Step-by-step explanation:

We would apply the straight line depreciation method. In this method, the value of the asset(bulldozer) is reduced linearly over its useful life until it reaches its salvage value. The formula is expressed as

Annual depreciation expense =

(Cost of the asset - salvage value)/useful life of the asset.

From the given information,

Useful life = 23 years

Salvage value of the bulldozer = $14950

Cost of the new bulldozer is $138000

Therefore

Annual depreciation = (138000 - 14950)/ 23 = $5350

The value of the bulldozer at any point would be V. Therefore

5350 = (138000 - V)/ t

5350t = 138000 - V

V = 138000 - 5350t

The value of the bulldozer after 3 years would be

V = 138000 - 5350×3 = $121950

Answer:

Earth's average distance from the Sun is [tex]375[/tex] times more than is from the Moon.

Step-by-step explanation:

Given the average distance of Earth from the Sun is [tex]1.5\times10^{8}[/tex] kilometers.

Also, the approximate distance of Earth from the Moon is [tex]4\times10^{5}[/tex] kilometers.

For this, we will divide the average distance of Earth from the Sun by approximate distance of Earth from the Moon.

So,

[tex]\frac{1.5\times10^{8}}{4\times10^{5}}=375\ km[/tex]

Answer:

There is a 9.75% probability that exactly 2 of them have type O blood.

Step-by-step explanation:

For each children, there is only two possible outcomes. Either they have blood type O, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

4 children, so [tex]n = 4[/tex]

Probability 0.15 of having blood type O, so [tex]p = 0.15[/tex]

If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.15)^{2}.(0.85)^{2} = 0.0975[/tex]

There is a 9.75% probability that exactly 2 of them have type O blood.

Answer:

MArginal productivity: [tex]\frac{dt}{dL}=-0.0002[/tex]

We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

Step-by-step explanation:

The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.

In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.

The marginal productivity can be expressed as:

[tex]\frac{dt}{dL}[/tex]

where dt is the variation in time and dL is the variation in training yards.

We can not derive the function because it is not defined, but we can approximate with the last two points given:

[tex]\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002[/tex]

Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

This is an approximation that is valid in the interval of 60,000 to  70,000 yards of training.

The marginal productivity of the number of yards Ben practices is 0.00051 seconds per yard.

The marginal productivity of the number of yards Ben practices can be calculated by dividing the change in his performance (measured in seconds) by the change in the number of yards he practices per week.

To find the marginal productivity between 50,000 and 60,000 yards per week, subtract Ben's initial time of 51.5 seconds from his new time of 46.4 seconds. This gives us a change in time of 5.1 seconds. Then divide the change in time by the change in yards, which is 10,000 yards (60,000 - 50,000).

Marginal productivity = Change in time / Change in yards = 5.1 seconds / 10,000 yards = 0.00051 seconds per yard.

https://brainly.com/question/34868070

#SPJ3

Answer:

Step-by-step explanation:

Given

There are 45 numbers out of which 5 numbers are required to win an enormous sum of money

No of ways  in which 5 numbers can be selected out of 45 numbers.

[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]

here n=45, r=5

[tex]^{45}C_5=\frac{45!}{40!5!}[/tex]

[tex]^{45}C_5=1221759[/tex]

out of which there is only combination foe contest winning

[tex]P=\frac{1}{1221759}[/tex]

When 100 different tickets are bought then

Probability of winning[tex]=\frac{100}{1221759}[/tex]

The probability of winning with one combination of five numbers is [tex]\( \frac{1}{1,221,759} \)[/tex] .

The probability of winning if 100 different lottery tickets are purchased is [tex]\( 1 - \left(1 - \frac{1}{1,221,759}\right)^{100} \)[/tex], approximately [tex]\( 0.0000182 \)[/tex].

Step 1

The probability of winning the lottery with one combination of five numbers can be calculated using the formula for combinations:

[tex]\[ P(\text{win with one ticket}) = \frac{1}{\binom{45}{5}} \][/tex]   Where [tex]\( \binom{45}{5} \)[/tex]represents the number of ways to choose 5 numbers from 45 without regard to the order. Calculating [tex]\( \binom{45}{5} \)[/tex] :

[tex]\[ \binom{45}{5} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759 \][/tex]

So, the probability [tex]\( P(\text{win with one ticket}) \)[/tex] is:

[tex]\[ P(\text{win with one ticket}) = \frac{1}{1,221,759} \approx 8.19 \times 10^{-7} \][/tex]

If 100 different lottery tickets are purchased, the probability of winning at least once is calculated using the complement rule:

[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{\binom{45}{5}} \right)^{100} \][/tex]

Step 2

Substituting [tex]\( \binom{45}{5} \)[/tex] :

[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{1,221,759} \right)^{100} \][/tex]

[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (1 - 8.19 \times 10^{-7})^{100} \][/tex]

[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (0.999999181)^{100} \][/tex]

[tex]\[ P(\text{win with 100 tickets}) \approx 1 - 0.9999818 \][/tex]

[tex]\[ P(\text{win with 100 tickets}) \approx 0.0000182 \][/tex]

The probability of winning the lottery with one ticket is approximately [tex]\( 8.19 \times 10^{-7} \)[/tex], while the probability of winning if 100 different tickets are purchased increases to approximately 0.0000182. Purchasing more tickets improves the chances of winning, but it remains a very low probability event due to the large number of possible combinations.

Answer:

The three statements are true.

Step-by-step explanation:

The question is incomplete:

A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.

A. There are 54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

C. There are 16 clients who own cats but not dogs.

A. There are 54 clients who own dogs.

TRUE. Of the 70 clients, only 36 own cats. There are left 34 clients that own only dogs. If we add the 20 clients that own both cats and dogs, we have 34+20=54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

TRUE. Of the 70 clients, only 36 own cats. Then, there are left 70-36=34 clients that own only dogs.

C. There are 16 clients who own cats but not dogs.

TRUE. Out of the 36 clients that own cats (only of with dogs), there are 20 that own both. Therefore, there are 36-20=16 clients that own only cats.

H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0 is the competing hypotheses to test the claim that there is an increase in the proportion of people who marry outside their race or ethnicity. The correct option is option A.

A hypothesis is a put forth explanation of a phenomenon (plural: hypotheses). A hypothesis must be testable according to the scientific method for it to be considered a scientific hypothesis. Scientific hypotheses are typically based on prior observations that cannot be adequately explained by the current body of knowledge.

H0:

p1 − p2 = 0;

HA: p1 −

p2 ≠ 0

H0:

p1 − p2 ≥ 0;

HA: p1 −

p2 < 0

H0:

p1 − p2 ≤ 0;

HA: p1 −

p2 > 0

H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0

Therefore, the correct option is option A.

To know more about hypothesis, here:

https://brainly.com/question/18064632

#SPJ1

Answer:

(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.

Step-by-step explanation:

Type I error is rejecting the true null hypothesis and type II error is not rejecting the false null hypothesis. Hence in this scenario, it will be:

A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.

Option C is correct.

Answer:

The answer is option A.

Step-by-step explanation:

If the report is made once a month, the daily variations (including the tendency to decrease efficiency on Friday) will be masked within the monthly result.

It would only generate an ethical problem in the case that Fridays fall in different samples, but in the case of the monthly report there are usually 4 or 5 Fridays included in each sample.

One way to do this is to recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]

so that

[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]

(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.

Answer: the cost of one hosta is $7

the cost of one geramium is $12

Step-by-step explanation:

Let x represent the cost of one hosta.

Let y represent the cost of one geramium.

They bought their supplies from the same store. Jill spent $101 on 11 hostas and 2 geraniums. This means that

11x + 2y = 101 - - - - - - - -1

Kim spent $80 on 8 hostas and 2 geraniums. This means that

8x + 2y = 80 - - - - - - - - -2

We will eliminate y by subtracting equation 2 from equation 1, it becomes

3x = 21

x = 21/3 = 7

Substituting x = 7 into equation 2, it becomes

8 × 7 + 2y = 80

56 + 2y = 80

2y = 80 - 56 = 24

y = 24/2 = 12

Answer: d) 6,534,385

Therefore, the number of ways of selecting 9 hearts is 6,534,385

Step-by-step explanation:

Given;

Number of cards to be chosen = 12

Number of hearts to be chosen = 9

Number of non-hearts to be selected = 3

Total number of cards = 52

Number of hearts total = 13

Number of non-hearts total= 39

The number of ways of selecting hearts can be given by the illustration below.

Number of ways of selecting hearts × number of ways of selecting non-hearts

N = 13C9 × 39C3 (it's combination since order is not important)

N = 13!/(9! ×4!) × 39!/(36! × 3!)

N = 6,534,385

Therefore, the number of ways of selecting 9 hearts is 6,534,385

The number of ways to choose 9 hearts from 12 cards is 220, option (b) is correct.

To solve this problem, we'll use the combination formula, which is used when selecting objects from a larger set without regard to the order. The formula for combinations is:

[tex]\[ \text{C}(n, k) = \frac{n!}{k!(n - k)!} \][/tex]

where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and (!) denotes factorial.

Given that we need to choose 12 cards from a deck of 52, and we want to know the number of ways to choose 9 hearts, which is [tex]\( k = 9 \)[/tex] hearts out of a total of [tex]\( n = 12 \)[/tex] cards.

Step 1:

Calculate the number of ways to choose 9 hearts from the 12 chosen cards:

[tex]\[ \text{C}(12, 9) = \frac{12!}{9!(12 - 9)!} \][/tex]

Step 2:

Simplify the expression:

[tex]\[ \text{C}(12, 9) = \frac{12!}{9!3!} \][/tex]

[tex]\[ \text{C}(12, 9) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \][/tex]

[tex]\[ \text{C}(12, 9) = \frac{1320}{6} \][/tex]

[tex]\[ \text{C}(12, 9) = 220 \][/tex]

Therefore, there are 220 ways to choose 9 hearts from 12 cards. So, the correct answer is option (b).

Answer:

e)  There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in

Step-by-step explanation:

Given that an ecologist collects data on the weights and basal trunk diameters of a randomly selected sample of felled trees. The ecologist fits a simple linear regression model, predicting tree weight from the trunk's cross-sectional area. After ensuring that all assumptions of the linear model are met, the ecologist computes a 95% confidence interval for the predicted mean weight of trees with a trunk cross-sectional area of 452 inches

We can interpret the confidence interval as

we are 95% confident that for samples of trees largely drawn with 452 inches the mean of these trees would fall within this interval.

e)  There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in

Answer:

We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, [tex]\bar{x}[/tex] = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 101.5\\H_A: \mu > 101.5[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.96[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Answer:

Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]

[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]

b) 2.70

[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]

b. 1.85

[tex]p_v =P(Z>1.85)=0.032[/tex]

b. 0.03

a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=164[/tex] represent the mean for the sample 1

[tex]\bar X_{2}=159[/tex] represent the mean for the sample 2

[tex]\sigma_{1}=12.5[/tex] represent the population standard deviation for the sample 1

[tex]s_{2}=9.25[/tex] represent the population standard deviation for the sample B2

[tex]n_{1}=35[/tex] sample size selected 1

[tex]n_{2}=30[/tex] sample size selected 2

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Standard error

The standard error on this case is given by:

[tex]SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]

Replacing the values that we have we got:

[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]

b. 2.70

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]  

P-value

Since is a one side right tailed test the p value would be:

[tex]p_v =P(Z>1.85)=0.032[/tex]

b. 0.03

Conclusion

If we compare the p value and the significance level given [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.

a. We can reject H(0) in favor of H(a)

Answer:

[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]

Step-by-step explanation:

probability distribution fucntion is given as

[tex] F_x = P(X \leq x)[/tex]

       [tex] = \frac{x -a}{b -a}      a<x< b[/tex]

where a indicate lower end estimate = $250,000

b indicate high end estimate = $350,000

probability greater than $337500

[tex] P(X \geq 337,500) = 1- P(X < 337500)[/tex]

                               [tex] = 1 - \frac{x -a}{b -a}[/tex]

                               [tex] = 1 - \frac{337500 - 250000}{350000 - 250000}[/tex]

[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]

Calculate the probability of net income being greater than or equal to $337,500 for Excellence Corporation following a continuous uniform distribution. Probability will be equal to 12.5%.

The probability that net income will be greater than or equal to $337,500 is 0.25. To calculate this probability for a continuous uniform distribution, we need to find the proportion of the area under the distribution curve that corresponds to values greater than $337,500.

Step-by-step calculation:

Calculate the total range of net income: $350,000 - $250,000 = $100,000

Calculate the proportion of the area for values greater than $337,500: ($350,000 - $337,500) / $100,000 = 0.125

Since the distribution is continuous and uniform, the probability is equal to the proportion calculated: 0.125 = 12.5%

Answer:

Cyril's adjusted gross income is $12000.

Step-by-step explanation:

First, we have that the adjusted gross income is given by the sum of the income that is not taxable.  

For our case, we have the following data:

Social security:  12000 (This is not taxable)

Wages:  5000 (This is taxable)

Interest and dividends:  4000 (This is taxable)

Unemployment compensation:  3000 (This is taxable)

municipal bond interest: $1,500 (This is not taxable)

Then, adding up the income that's deductible, we have that Cyril's adjusted gross income is:

AGI=5000+4000+3000=$12000

Answer:

  Distribute 2 to (x + 6) and 3 to (x – 4)

Step-by-step explanation:

A good first step for an equation such as this is to eliminate parentheses. That is accomplished by using the distributive property on the left and right. That property requires each term inside parentheses be multiplied by the factor outside parentheses. The process can be described as ...

  Distribute 2 to (x + 6) and 3 to (x – 4)

_____

Comment on further steps

The result of the first step above would be ...

  2x + 12 = 3x - 12 + 5

  19 = x . . . . . . . . . . . next step: add 7-2x to both sides and collect terms

_____

Comment on answer wording

The last answer choice uses the wording "move the 6 from the left side of the equation to the right side". This sort of description is often invoked when a constant needs to be removed from one side of the equation or the other.

  There is no property of equality that supports this action.

If you want to "move" a constant, you can invoke the addition property of equality that lets you add the same number to both sides of the equation. The number you would choose to add is the additive inverse of the one you want to "move."

In the above solution we have the sum -12+5 on the right after the first step. We want to eliminate that sum, so in the next step we show adding the additive inverse of the value of that sum. The sum of -12+5 is -7, and we add 7 to make that sum into the additive identity element, 0. (Likewise, we add -2x to turn the term 2x into the additive identity element, 0.) After these additions, we have ...

  2x -2x +12 +7 = 3x -2x -12 +5 +7

  0 + 19 = x + 0

  19 = x

The relevant property of equality is the addition property of equality, which says I can add the same amount to both sides of the equation. I can't "move" anything without violating the equal sign, but I can use the addition and multiplication properties of equality to do the same thing to both sides of the equation.

Answer:

The correct answer is Distribute 2 to (x + 6) and 3 to (x – 4).

Step-by-step explanation:

Answer:

The amount fertilizer A the farmer should use while still meeting the minimum requirement is approximately seventy-six 50-lb bags which would cost $6080.

The amount of fertilizer B the farmer should use while still meeting the minimum requirement is approximately seventy-one 50- lb bags which would cost $2120

Step-by-step explanation:

Fertilizer A

Let x represent the number of 50- lb bags needed for the minimum requirement

Cost of I 50-lb bag = $80

Cost of x 50-lb bag = $80x

1 50-lb bag contains 14lb nutrient ( 8lb Nitrogen+ 2lb Phosphorus + 4lb Potassium)

x 50-lb bag contains 1060lb nutrient (440lb N + 260lb P + 360lb K) which is the minimum requirement

x = 1060lb ÷ 14lb = 75.71

x = 76 ( to the nearest whole number)

76 50-lb bags would cost 76×$80 = $6080

Fertilizer B

Cost of a 50- lb bag is $30

Let y be the number of bags required

Cost of y bag is $30y

y = 1060÷ 15= 71

Cost = 71× $30 = $2130

Answer:

0 fertilizer A

88 fertilizer B

Step-by-step explanation:

Shown in the attachment

Answer:

Option A.

Step-by-step explanation:

The given probability table is

Number of cars X: 0       1         2        3        4        5      6      7      8

Probability :0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001

It is given that a housing company builds houses with two car garages.

We need to find the percent of households have more cars than the garage can hold.

[tex]P(X>2)=1-P(X\leq 2)[/tex]

[tex]P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]

Substitute the probability values from the given table.

[tex]P(X>2)=1-[0.087+0.323+0.363][/tex]

[tex]P(X>2)=1-0.773[/tex]

[tex]P(X>2)=0.227[/tex]

It means 22.7% of households have more cars than the garage can hold.

Therefore, the correct option is A.

The percent of households that have more cars (including SUVs and light trucks) than a two-car garage can accommodate is approximately 22.7%.

To answer the student's question, let's examine the probability model provided. The random variable X represents the number of cars, including SUVs and light trucks, in a randomly selected American household.

If a housing company builds houses with two-car garages, we want to determine what percent of households have more cars than the garage can hold. To find this, we would add up the probabilities for the households that own more than 2 cars.

The probabilities for owning 3, 4, 5, 6, 7 or 8 cars are 0.144, 0.053, 0.019, 0.007, 0.002, and 0.001, respectively. Adding all these probabilities gives us:

0.144 + 0.053 + 0.019 + 0.007 + 0.002 + 0.001 = 0.226

To convert this to a percentage, we multiply by 100, which gives us 22.6%. Therefore, approximately 22.7% of households have more cars than the garage can hold, which corresponds to answer choice A.

https://brainly.com/question/32117953

#SPJ3