Which graph shows the solution set for Negative 4.4 greater-than-or-equal-to 1.6 x minus 3.6?
A number line going from negative 3 to positive 3. A closed circle is at negative 0.5. Everything to the left of the circle is shaded.
A number line going from negative 3 to positive 3. A closed circle is at negative 0.5. Everything to the right of the circle is shaded.
A number line going from negative 7 to negative 1. A closed circle is at negative 5. Everything to the left of the circle is shaded.
A number line going from negative 7 to negative 1. A closed circle is at negative 5. Everything to the right of the circle is shaded.

Answer:

Step-by-step explanation:

Given data

Average sales = 8000

n = 64

standard deviation = 1200

      8300

The solution is attached in the picture below

Answer:

We conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.

Step-by-step explanation:

We are given that an article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees.

You conduct a survey of 39 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2.

Let [tex]\mu[/tex] = average time taken to finish their undergraduate degrees.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 4.5 years     {means that the mean time taken to finish undergraduate degrees is equal to 4.5 years}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 4.5 years     {means that the mean time taken to finish undergraduate degrees is longer than 4.5 years}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                           T.S. =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean time = 5.1 years

            s = sample standard deviation = 1.2 years

            n = sample of students = 39

So, the test statistics  =  [tex]\frac{5.1-4.5}{\frac{1.2}{\sqrt{39} } }[/tex]  ~ [tex]t_3_8[/tex]

                                     =  3.122

The value of t test statistics is 3.122.

Now, at 1% significance level the t table gives critical value of 2.429 at 38 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.122 > 2.429, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.

Ok, what's the full question here

Answer:

B. 104

Step-by-step explanation:

Just find the area of the wall and subtract the area of the window. The area of the wall is 10 times 11 which is 110. The area of the window is 2 times 3 which is 6. 110 minus 6 is 104.

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size  'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample  S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size  n₂ = 24

mean of the second sample  "x₂⁻"= 14.89 pounds

standard deviation of second sample  S₂ =  2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

   [tex]t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }[/tex]

   where         [tex]S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}[/tex]

                      [tex]S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}[/tex]

             substitute values and we get  S² =  40.988

    [tex]t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}[/tex]

    t =  1.3622

  Calculated value t = 1.3622

Tabulated value 't' =  2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

Answer:

[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]    

The degrees of freedom are given by:

[tex] df= n-1 = 4934-1= 4933[/tex]

Then the p value for this case calculated as:

[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]  

Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion

Step-by-step explanation:

Information provided

[tex]\bar X=27.1[/tex] represent the sample mean

[tex]s=7.3[/tex] represent the sample standard deviation

[tex]n=4934[/tex] sample size  

[tex]\mu_o =27.4[/tex] represent the value to test

t would represent the statistic  

[tex]p_v[/tex] represent the p value

System of hypothesis

We want to verify if the mean age of​ full-time students did decline (less than 27.4), the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 27.4[/tex]  

Alternative hypothesis:[tex]\mu < 27.4[/tex]  

The statistic for this case is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Replacing the info we got:

[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]    

The degrees of freedom are given by:

[tex] df= n-1 = 4934-1= 4933[/tex]

Then the p value for this case calculated as:

[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]  

Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion

The combined weight of the two kittens is 70 ounces.

To find the combined weight of the two kittens, we'll start by converting the weight of the first kitten to ounces.

1 pound is equal to 16 ounces, so 2 pounds is equal to 2 x 16 = 32 ounces.

Therefore, the first kitten weighs 32 + 4 = 36 ounces.

The weight of the second kitten is 2 ounces less, so we subtract 2 from the weight of the first kitten: 36 - 2 = 34 ounces.

Finally, we can find the combined weight by adding the weights of the two kittens together: 36 + 34 = 70 ounces.

Therefore, the combined weight of the two kittens is 70 ounces.

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Answer:

By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed(bell-shaped) 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 = 30.25

Standard deviation = 12.60

Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85

17.65 = 30.25 - 1*12.60

So 17.65 is one standard deviation below the mean.

42.85 = 30.25 + 1*12.60

So 42.85 is one standard deviation above the mean

By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85

Answer:

6.94% probability that fewer than 35 use self-directed work teams as a management tool

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 70, p = 0.58[/tex]

So

[tex]\mu = E(X) = np = 70*0.58 = 40.6[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{70*0.58*0.42} = 4.13[/tex]

What is the probability that fewer than 35 use self-directed work teams as a management tool?

Using continuity correction, this is P(X < 35 - 0.5) = P(X < 34.5), which is the pvalue of Z when X = 34.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{34.5 - 40.6}{4.13}[/tex]

[tex]Z = -1.48[/tex]

[tex]Z = -1.48[/tex] has a pvalue of 0.0694.

6.94% probability that fewer than 35 use self-directed work teams as a management tool

The question requires calculation of binomial probability. Given that the rate of success is 58% (or 0.58) and we're trying to find the likelihood of fewer than 35 successes out of 70 trials, one must sum the binomial probabilities from k=0 to 34.

This problem is about calculation of binomial probability, which is a specific type of probability that deals with experiments that have two possible outcomes: success (in this case, using self-directed work teams) or failure (not using self-directed work teams). Given that the rate of success is 58% (or 0.58 as a decimal), and we're looking for the probability of fewer than 35 successes out of 70 trials (or managers), we can solve using the formula for binomial probability.

The basic form of the binomial formula is: [tex]P(X=k) = C(n, k) * (p^k) * ((1-p)^{(n-k)})[/tex], where n is the number of trials, k is the number of successful trials, p is the probability of success, and C(n, k) is a combination which calculates the number of possible combinations of n items taken k at a time.

To find P(X<35), we sum from k=0 to 34. Thus, this involves a fair amount of calculation, and you may want to use software or a calculator that has binomial probability functionality.

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Complete question :

You have decided to stop drinking Starbucks coffee and invest that money in an IRA. If you deposit $492 each month earning 6% interest, how much will you have in the account after 40 years?

Answer:

$250,329.60

Step-by-step explanation:

Given:

Principal, P= $492 per month

= $492 * 12 = $5,904 per year

Rate, R= 6%

Time, T=  40 years

Let's first find the amount after 40 years.

Amount = 5940 * 40 = $236,160

The interest after 40 years =

[tex] Interest = \frac{PRT}{100} = \frac{5904 * 6 * 40}{100}[/tex]

Interest = $14,169.60

The total amount I will have in my account after 40 years:

Amount + interest

= $236,160 + $14,169.60

= $250,329.60

Answer:

99

Step-by-step explanation:

11 crows : 3 hawks

363 crows: X hawks

X/363 = 3/11

X = 363 × 3/11

X = 99

Answer: The upper control limit is 1.40581

The lower control limit is 0.000

Explanation: Check attachment

The upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.

To compute the upper and lower control limits for the R chart, we need to calculate the average range (R-bar) and the control limits.

First, let's calculate the range (R) for each sample by subtracting the smallest value from the largest value:

Sample 1: 1.0

Sample 2: 0.9

Sample 3: 0.9

Sample 4: 0.4

Sample 5: 0.5

Sample 6: 1.1

Sample 7: 0.9

Sample 8: 0.3

Sample 9: 0.2

Sample 10: 0.6

Sample 11: 0.6

Sample 12: 0.2

Sample 13: 1.3

Sample 14: 0.6

Sample 15: 0.8

Sample 16: 1.1

Sample 17: 0.6

Sample 18: 0.5

Sample 19: 0.4

Sample 20: 0.6

Next, calculate the average range (R-bar) by summing up all the ranges and dividing by the number of samples:

R-bar = (1.0 + 0.9 + 0.9 + 0.4 + 0.5 + 1.1 + 0.9 + 0.3 + 0.2 + 0.6 + 0.6 + 0.2 + 1.3 + 0.6 + 0.8 + 1.1 + 0.6 + 0.5 + 0.4 + 0.6) / 20

R-bar = 0.735

To calculate the upper control limit (UCL) for the R chart, multiply the R-bar by a constant factor. For small sample sizes like this (n=5), the constant factor is typically 2.704:

UCL = R-bar * 2.704

UCL = 0.735 * 2.704

UCL = 1.986

Finally, to calculate the lower control limit (LCL) for the R chart, multiply the R-bar by another constant factor. For small sample sizes like this, the constant factor is typically 0:

LCL = R-bar * 0

LCL = 0

Therefore, the upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.

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Answer:

200.96 in^2

Step-by-step explanation:

We want to find the area

A = pi r^2

A = pi (8)^2

A = 3.14 (64)

A =200.96

Answer: yes, GM is expected to provide performance greater than 123

Step-by-step explanation:

b) It is expected that the GM achieves a performance greater than 123. This situation occurs because 123 is less than the average for GM but is greater than the average for Regular. Thus we observe that P (performance> 123 | GM)> 0.5 and P (performance> 123 | Regular) <0.5

Answer:

[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=23-1=22[/tex]  

And the p value taking in count that we have a bilateral test we got:

[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

Step-by-step explanation:

Information given

[tex]\bar X=435[/tex] represent the mean for the weight

[tex]s=\sqrt{841}=29[/tex] represent the sample standard deviation

[tex]n=23[/tex] sample size  

[tex]\mu_o =430[/tex] represent the value that we want to verify

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value

System of hypothesis

We are trying to proof if the filling machine works correctly at the 430 gram setting, so then the system of hypothesis for this case are:

Null hypothesis:[tex]\mu = 430[/tex]  

Alternative hypothesis:[tex]\mu \neq 430[/tex]  

In order to cehck the hypothesis the statistic for a one sample mean test is given by

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Replacing the info given we have this:

[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=23-1=22[/tex]  

And the p value taking in count that we have a bilateral test we got:

[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

The type of hypothesis test used in this scenario is a two-tailed t-test for a single sample mean.

Step 1: State the hypotheses

- Null hypothesis H₀: The mean weight of the bags is 430 grams [tex]\mu = 430[/tex]

- Alternative hypothesis H₁: The mean weight of the bags is not 430 grams [tex]\mu \neq 430[/tex]

Step 2: Select the significance level

- The level of significance [tex]\alpha[/tex] is 0.05.

Step 3: Calculate the test statistic

- Sample mean [tex]\bar{x}\)[/tex] = 435 grams

- Population mean [tex]\mu\)[/tex] = 430 grams

- Sample size n = 23

- Sample variance s² = 841

- Sample standard deviation s = √841 = 29

The test statistic (t) is calculated using the formula:

[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]

[tex]\[ t = \frac{435 - 430}{29 / \sqrt{23}} \][/tex]

[tex]\[ t = \frac{5}{6.048} \][/tex]

t ≈ 0.8266

Step 4: Determine the degrees of freedom and critical value

- Degrees of freedom [tex](\(df\)) = \(n - 1[/tex]

= 23 - 1

= 22

- For a two-tailed test at [tex]\(\alpha = 0.05\)[/tex] with 22 degrees of freedom, the critical t-values are approximately [tex]\(\pm 2.074[/tex].

Step 5: Make a decision

- Compare the calculated test statistic 0.8266 with the critical t-values [tex]\pm 2.074[/tex].

- If the test statistic is within the range -2.074 to 2.074, fail to reject the null hypothesis.

- If the test statistic is outside this range, reject the null hypothesis.

Since 0.8266 is within the range -2.074 to 2.074, we fail to reject the null hypothesis.

Conclusion:

- There is not enough evidence to suggest that the mean weight of the bags is different from 430 grams at the 0.05 significance level.

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

After performing the calculations, the right matrix multiplication by D changes each column of A and the left matrix multiplication by D changes each row of A. Also, there are an infinite number of 3x3 matrices that satisfy the condition AB ≠ BA

To compute the products AD and DA, we first need to understand how matrix multiplication works. Given matrices A and D as presented, it is important to understand that when A is multiplied by D on the right side (AD), each column of A is multiplied by the corresponding diagonal entry of D. Conversely, when A is multiplied by D on the left side (DA), each row of A is multiplied by the corresponding diagonal entry of D. This is true for any two matrices you want to multiply. Therefore, the correct answer to the first part of the question is statement D.

For the second part of the question, we're seeking a non-identity, non-zero 3x3 matrix B such that AB is not equal to BA. This is generally quite difficult and requires some trial and error. There are an infinite number of matrices B that satisfy the condition AB ≠ BA. The exact form of B depends on A and can be quite complex, but B isn't necessarily unique. Therefore, the correct answer to this question is B: There are infinitely many solutions.

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Answer:

[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=22-1=21[/tex]  

The p value is given by:

[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]  

The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600

Step-by-step explanation:

Information given

[tex]\bar X=33400[/tex] represent the sample mean

[tex]s=1520[/tex] represent the sample standard deviation

[tex]n=22[/tex] sample size  

[tex]\mu_o =33600[/tex] represent the value that we want to analyze

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value for the test

System of hypothesis

We want to check if the true mean is lower than 33600, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 33600[/tex]  

Alternative hypothesis:[tex]\mu < 33600[/tex]  

The statistic is given:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Replacing the data given we got:

[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=22-1=21[/tex]  

The p value is given by:

[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]  

The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600

Answer:

15-25% of survivors were incapacitated

Step-by-step explanation:

According to the analysis conducted by Norris and her colleague in 2006, it was found that out of the 160 empirical studies conducted, 41% showed evidence of severe to very severe impairment (interference with functioning) among disaster survivors.

This in turn corresponds to 15-25% increase in demand for mental health services by the population affected by this disasters.

This percentage of people affected by disasters suffer from disaster's syndrome which include but not limited to shock, bewilderment and void of deep emotion.

Hence they become incapacitated.

The question is incomplete, as the required details are not given. However, I will give a general explanation on how to determine percentages.

To calculate the percentage of survivors that were incapacitated, we need:

Assume that:

[tex]n = 250[/tex] --- survivors

[tex]k = 100[/tex] ---  survivors that were incapacitated

The percentage (p) is calculated as follows:

[tex]p = \frac kn \times 100\%[/tex]

So, we have:

[tex]p = \frac{100}{250} \times 100\%[/tex]

[tex]p = \frac{100\times 100}{250} \%[/tex]

[tex]p = \frac{10000}{250} \%[/tex]

[tex]p = 40 \%[/tex]

Using the assumed values, 40% of the survivors were incapacitated.

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Answer:

8000

Step-by-step explanation:

1/4 of 16000 is 4000

16000-4000=12000 1/3 of 12000 is also 4000

12000-4000=8000

Answer: 1007.28

Step-by-step explanation:

Given : The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with

[tex]\mu=820\ \ \ ,\ \sigma=200[/tex]

If a college requires a student to be in the top 15 % of students taking this test, it means that they want the students that score 85 percentile or above.

Let X be the scores of any random student, we require

[tex]P(X<x)=0.85[/tex], where x is minimum score that such a student can obtain and still qualify for admission at the college.

Formula for z-score = [tex]z=\frac{x-\mu}{\sigma}=\frac{x-820}{200}[/tex]  ...(i)

From normal z-value table , [tex]P(z<1.036)=0.85[/tex]...(ii)

From (i) and (ii) , we get

[tex]\frac{x-820}{200}=1.0364\\\\\Rightarrow\ x-820=207.28\\\\\Rightarrow\ x=207.28+820=1007.28[/tex]

Hence, the minimum score that such a student can obtain and still qualify for admission at the college is 1007.28.

Answer:

Step 2

Step-by-step explanation:

The quadratic formula is given by [tex]x=-b+-\frac{\sqrt{b^{2}-4ac } }{2a}[/tex]

Our equation is y = 2x²-x-6

So here our a = 2, b = -1, and c = -6

We can now plug these numbers into our formula

[tex]x= -(-1) +-\frac{\sqrt{(-1)^{2}-4(2)(-6) } }{2(2)} = 1 +-\frac{\sqrt{1+24} }{4} = 1+-\frac{\sqrt{25} }{4}[/tex]

Step 2 is incorrect because it states that "x equals the negative of negative 1 plus or minus the square root of negative one plus forty-eight all over two times 2."

The correct statement would be "x equals the negative of negative 1 plus or minus the square root of positive one plus forty-eight all over two times 2.", because the square of a negative is positive, resulting in a positive one.

Since this step is incorrect, the steps after are also incorrect, but Harry went wrong at Step 2

Answer:

The above answer is correct.

Step-by-step explanation:

I got it right on the test

Answer:

12.36

Step-by-step explanation:

Answer:

  f)  a[n] = -(-2)^n +2^n

  g)  a[n] = (1/2)((-2)^-n +2^-n)

Step-by-step explanation:

Both of these problems are solved in the same way. The characteristic equation comes from ...

  a[n] -k²·a[n-2] = 0

Using a[n] = r^n, we have ...

  r^n -k²r^(n-2) = 0

  r^(n-2)(r² -k²) = 0

  r² -k² = 0

  r = ±k

  a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q

We find p and q from the initial conditions.

__

f) k² = 4, so k = 2.

  a[0] = 0 = p + q

  a[1] = 4 = -2p +2q

Dividing the second equation by 2 and adding the first, we have ...

  2 = 2q

  q = 1

  p = -1

The solution is a[n] = -(-2)^n +2^n.

__

g) k² = 1/4, so k = 1/2.

  a[0] = 1 = p + q

  a[1] = 0 = -p/2 +q/2

Multiplying the first equation by 1/2 and adding the second, we get ...

  1/2 = q

  p = 1 -q = 1/2

Using k = 2^-1, we can write the solution as follows.

The solution is a[n] = (1/2)((-2)^-n +2^-n).

Answer:

10 (3x+7)

Step-by-step explanation:

factor out 10

10 (3x+7)

Answer:

  (-5, 0) ∪ (5, ∞)

Step-by-step explanation:

I find a graph convenient for this purpose. (See below)

__

When you want to find where a function is increasing or decreasing, you want to look at the sign of the derivative. Here, the derivative is ...

  f'(x) = 4x^3 -100x = 4x(x^2 -25) = 4x(x +5)(x -5)

This has zeros at x=-5, x=0, and x=5. The sign of the derivative will be positive when 0 or 2 factors have negative signs. The signs change at the zeros. So, the intervals of f' having a positive sign are (-5, 0) and (5, ∞).

The details provided do not correspond to a coherent mathematics problem regarding the function [tex]f(x) = x^4 - 50x^2 + 3[/tex]. Therefore, an accurate response cannot be provided without further information or correct context.

To address the question about the function [tex]f(x) = x^4 - 50x^2 + 3[/tex], we need to analyze its intervals of increase and decrease, as well as find any local extrema and points of concavity. However, the question as posed does not provide enough context or coherent detail for the actions requested, such as finding the inflection points or intervals of concavity, since no specific function was clearly defined. Instead, various unrelated Mathematics problems are listed, each of which is missing comprehensive details needed to provide an accurate answer.

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Answer: I think c,e

Step-by-step explanation:Sorry if it wrong

Answer:

C

Step-by-step explanation:

IM DORA

Answer:

Step-by-step explanation:

Answer:

10.5m/h

Step-by-step explanation:

speed at 40 revolutions= 7/60 m/minute = 0.1167

circumference of wheel = 101167/40 = 7/2400

now,

at 60 pedals = speed = 60*7/2400 = 0.175m/minute

speed= 0.175*60 m/h = 10.5

Answer:

Step-by-step explanation:

This is a test of a single population mean since we are dealing with mean.

From the information given,

Null hypothesis is expressed as

H0:μ=1.5

The alternative hypothesis is expressed as

Ha:μ>1.5

This is a right tailed test

The decision rule is to reject the null hypothesis if the significance level is greater than the p value and accept the null hypothesis if the significance level is less than the p value.

p value = 0.052

Significance level, α = 0.05

Since α = 0.05 < p = 0.052, the true statement would be

At the α=0.05 significance level, you have proven that H0 is true. B.

Final answer:

The P-value of 0.052 provides some evidence against the null hypothesis H0, although not enough to reject it at the α = 0.05 significance level. A study with a larger sample size may be worthwhile to investigate the true mean nicotine content further.

Explanation:

With a P-value of 0.052 for the hypothesis test, comparing it against a significance level alpha (α) = 0.05 determines whether the null hypothesis (H0) is rejected or not. Despite the P-value being slightly above the significance level, the correct interpretation is Option C: 'There is some evidence against H0, and a study using a larger sample size may be worthwhile.' This suggests that while there isn't enough evidence to reject the null hypothesis at α = 0.05, the result is close enough to warrant further investigation.

We cannot claim that the null hypothesis has been proven true as statistical hypothesis testing never proves a hypothesis, but only provides evidence against it (Option A is incorrect). The data suggests that further investigation could be useful, instead of being unfruitful (Option B is an incorrect view). Lastly, there is evidence pointing towards the alternative hypothesis (Ha), it's not that there's no evidence at all (Option D is incorrect).