Let U represent the set of people in a certain community who were asked if they subscribe to an information source.
Let D ={ x ∈ U | x subscribes to The Daily Informer }
and
N ={ x ∈ U | x subscribes to News Magazine }


(Assume these sets are not disjoint.) Write the set that represents the set of people surveyed who subscribe to exactly one of the two news sources given.

a) N ∪ D
b) ( N c ∩ D ) ∪ ( N ∩ D c )
c) N c ∩ D
d) ( N ∩ D )c
e) ( N c ∪ D ) ∩ ( N ∪ D c )

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Answer:

B. The true population percentage is likely to be within 5% of the sample percentage.

Step-by-step explanation:

These samples have a confidence level of x%, which leads us to a confidence interval with a margin of error is M%.

This means that we are x% sure that the true population percentage is within M% of the sample percentage.

In this problem, the true population percentage is likely to be within 5% of the sample percentage.

So the correct answer is:

B. The true population percentage is likely to be within 5% of the sample percentage.

Answer:

a) The 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

b)The 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

Step-by-step explanation:

1) Previous concepts

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=30.5[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=6.7 represent the sample standard deviation

n=235 represent the sample size  

2) First confidence interval

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=235-1=234[/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(0.025,234)".And we see that [tex]t_{\alpha/2}=1.97[/tex]

Now we have everything in order to replace into formula (1):

[tex]30.5-1.97\frac{6.7}{\sqrt{235}}=29.639[/tex]    

[tex]30.5+1.97\frac{6.7}{\sqrt{235}}=31.361[/tex]    

So on this case the 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

3) Second confidence interval

[tex]\bar X=30.4[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=2.3 represent the sample standard deviation

n=12 represent the sample size  

[tex]df=n-1=12-1=11[/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(0.025,11)".And we see that [tex]t_{\alpha/2}=2.20[/tex]

Now we have everything in order to replace into formula (1):

[tex]30.4-2.20\frac{2.3}{\sqrt{12}}=28.939[/tex]    

[tex]30.4+2.20\frac{2.3}{\sqrt{12}}=31.861[/tex]    

So on this case the 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

Answer:

110

Step-by-step explanation:

%20 increase = [tex]\frac{20}{100} 1300 = 260\\[/tex]

%50 decrease = [tex]\frac{50}{100}300 = 150[/tex]

%30 no change = [tex]\frac{30}{100}0 = 0[/tex]

Expected change = Increase - Decrease + No Change

Expected change = 260 - 150 + 0

Expected change = 110

Answer:

0% probability of a guest arriving by taxi and requesting parking for a car.

Step-by-step explanation:

Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.

In this problem, we have that a guest arriving via taxi and requesting parking at the hotel are mutually exclusive.

So there is a 0% probability of a guest arriving by taxi and requesting parking for a car.

Answer:

a) [tex]T(x)=300x+18x=318x[/tex]

b) [tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]

c) Yes.

Step-by-step explanation:

We have that:

[tex]Tires=$300\\Tax=0.006\\Discount=0.1\\[/tex]

And we have $10 free of taxes.

Making x= number of tires to buy, then we have that the total cost of tires is:

[tex]Total_{Tires}=300x[/tex]

So, what we pay for taxes is given by:

[tex]Taxes=(300x)(0.06)=18x[/tex]

a) Then, according to the above, we can write down the total cost before the discount as:

[tex]T(x)=300x+18x=318x[/tex]

b) And the total cost after discounts, is then given by:

[tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]

c) If the discount is added first, then less tax will be paid because the amount on which it is paid is lower. If the discount is added later, then the taxes will have been taxed on a higher amount, so it does matter whether they are added first or later.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, a hypothesis test needs to be conducted using the given sample data. The p-value needs to be calculated and compared to the significance level α to determine the conclusion. Without the p-value, we cannot make a definitive conclusion.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, the Department of Education would conduct a hypothesis test using the given sample data. The null hypothesis, denoted as H0, states that the average debt load is equal to $17,000, while the alternative hypothesis, denoted as Ha, states that the average debt load is not equal to $17,000.

Based on the given information, the sample mean debt load is $18,200, the population standard deviation is $4,200, and the sample size is 34. Using these values, we can calculate the p-value, which represents the probability of obtaining a sample mean as extreme as $18,200 or more extreme, assuming the null hypothesis is true. The p-value is the probability of observing a sample mean of $18,200 or more extreme, given that the average debt load is $17,000.

To determine the conclusion of the hypothesis test, we compare the p-value to the significance level α. In this case, the given α is 0.05. If the p-value is less than α, we reject the null hypothesis and conclude that the average debt load is not equal to $17,000. If the p-value is greater than or equal to α, we fail to reject the null hypothesis and cannot conclude that the average debt load is not equal to $17,000.

In this scenario, the p-value is not given, so we cannot make a definitive conclusion. We would need to calculate the p-value based on the given information to determine which statement is true.

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The correct option is A.  [tex]\text{Because the p-value is greater than}[/tex] [tex]\alpha,[/tex] [tex]\text{ we fail to reject the null hypothesis and conclude that the average debt load is equal to }[/tex] [tex]\$17,000.[/tex]

To determine the correct statement based on the hypothesis test conducted:

Given data:

Population standard deviation[tex](\( \sigma \)): \$4,200[/tex]

Sample size [tex](\( n \)): 34[/tex]

Sample mean [tex](\( \bar{x} \)): \$18,200[/tex]

Hypothesized population mean [tex](\( \mu_0 \))[/tex]: [tex]\$17,000.[/tex]

Significance level [tex](\alpha): 0.05[/tex]

Hypotheses

We are testing whether the average debt load [tex](\( \mu \))[/tex] of graduating students with a Bachelor's degree is equal to [tex]\$17,000.[/tex]

Null hypothesis [tex](\( H_0 \)): \( \mu = 17,000 \)[/tex]

Alternative hypothesis [tex](\( H_1 \)): \( \mu \neq 17,000 \)[/tex]

This is a two-tailed test because [tex]\( H_1 \)[/tex] states that [tex]\( \mu \)[/tex] is not equal to [tex]\$17,000.[/tex]

Test Statistic

Since the population standard deviation [tex](\( \sigma \))[/tex] is known and the sample size [tex](\( n \))[/tex] is greater than [tex]30[/tex], we use a z-test.

The test statistic [tex]\( z \)[/tex] is calculated as:

[tex]\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Calculate [tex]\( z \)[/tex]

[tex]\[ z = \frac{18,200 - 17,000}{\frac{4,200}{\sqrt{34}}} \][/tex]

[tex]\[ z = \frac{1,200}{\frac{4,200}{5.83}} \][/tex]

[tex]\[ z = \frac{1,200 \times 5.83}{4,200} \][/tex]

[tex]\[ z = 1.66 \][/tex]

P-Value Calculation

The p-value is the probability of observing a sample mean at least as extreme as [tex]18,200[/tex] if the null hypothesis [tex](\( \mu = 17,000 \))[/tex] is true. Since this is a two-tailed test, we consider both tails of the normal distribution.

From the standard normal distribution table:

The area to the right of [tex]\( z = 1.66 \)[/tex] (since [tex]\( z \)[/tex] is positive) is approximately [tex]0.0485.[/tex]

Therefore, the p-value for the two-tailed test is approximately [tex]\( 2 \times 0.0485 = 0.097 \).[/tex]

Conclusion

Compare the p-value [tex](0.097)[/tex] with the significance level [tex]\alpha = 0.05)\( \text{p-value} (0.097) > \alpha (0.05) \)[/tex]

Since the p-value is greater than the significance level ([tex]\alpha[/tex]), we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means we do not have enough evidence to conclude that the average debt load is different from [tex]\$17,000.[/tex]

Answer:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

[tex]\Chi^2_{crit} =10.851[/tex]

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

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

n = 21 sample size

[tex]s^2 =6.25[/tex] represent the sample variance

[tex]s=2.5[/tex] represent the sample deviation

[tex]\sigma^2_0 =16.81[/tex] represent the target variance

[tex]\sigma_o =4.1[/tex] the value that we want to test

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

t represent the statistic

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

State the null and alternative hypothesis

On this case we want to check if the population variance is lower (or if the deviation is lower than 4.1) than 16.81, so the system of hypothesis are:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

In order to check the hypothesis we need to calculate th statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 20 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,20)". And our critical value would be [tex]\Chi^2_{crit} =10.851[/tex]

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

What is the approximate p-value of the test?

For this case since we have a left tailed test the p value is given by:

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

Based on your answer for the p value, answer True or False as to whether the null hypothesis will be rejected.

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Answer:

Please read the answers below.

Step-by-step explanation:

Let's recall that in a square all its sides are equal length and all the four internal angles measure 90 °

5. If LN = 46, then we have:

OM = 46 (Same length than LN)

PN = LN/2 = 46/2 = 23

ON = √LN²/2 = √ 46²/2 = √ 2,116/2 = √ 1,058 = 32.53 (Rounding to two decimal places)

MN = ON = 32.53

6. m ∠EFG = 90°

m ∠GDH = ∠GDH/2 = 90/2 = 45°

m ∠FEG = ∠DEF/2 = 90/2 = 45°

m ∠DHG = 180 - (∠GDH + ∠DGH) = 180 - (45 + 45)= 180 - 90 = 90°

7. Solve for x

6x - 21 = ∠PQR/2

6x - 21 = 90/2

6x - 21 = 45

6x = 45 +21

6x = 66

x = 11

Answer:

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

1) 100 - 40 = 60

2) 800000 - 200000 = 600000

3) 300 + 80 = 410

4) 3000 + 800 = 1100

5) 600× 800 = 480000

6) 80×200 = 16000

7) 700×900 = 630000

8) 100×1000 = 100000

9) 800× 500 = 400000

10) 4000×80 = 320000

11) 800×100 = 80000

12) 800/20 = 40

13) 4000/70 = 57.1 = 60

14) 200/60 = 33.3 = 30

15) 600/80 = 7.5 = 8

16) 3000/200 = 15

17) 10000/200 = 50

18) 7000/300 = 23.3 = 20

19) 7000/400 = 17.5 = 18

20) 600000 / 2000 = 300

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

On demand

Step-by-step explanation:

On demand (fund Value Streams, not projects).

Lean Budgets is a collection of methods that significantly reduce overhead by financing and encouraging Value Streams instead of projects while preserving financial and user-fit governance. This is done by comprehensive work program assessment, active management of Epic finances, and complex budget changes.

Hence, the recommended frequency for updating lean budget distribution on demand that is fund value streams.

Lean budget distribution should be updated at least once per quarter, or as often as required by the specific circumstances of your organization. The goal of a lean budget is to optimize resource use and reduce waste to deliver more value to customers.

The recommended frequency for updating lean budget distribution is not fixed and largely depends on the specific needs, context, and circumstances of your organization. However, a common practice is to review and adjust the lean budget allocation at least once per quarter. This allows you to take into account any changes in strategy, project progress, and business environment. It's important to highlight that these revisions should be strategic and informed by data and not just random shifts in resource allocation. Remember, the main goal of a lean budget is to minimize waste and optimize resource utilization to bring more value to customers.

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At x = -2, there is a blue dot, which is a local minimum, because it is below the x axis.

The dot is located at y = -4

The answer is Yes there is a local minimum at x = -2

The local minimum is y = -4

Answer:

The answer is Yes there is a local minimum at x = -2

The local minimum is y = -4

Step-by-step explanation:

Found answer online so no work!!!

Answer:

Slope: 3

Y intercept: y = 75

Step-by-step explanation:

The slope of a linear function is the coefficient in front of x. In this case, 3 is being multiplied by x, indicating the slope is 3.

The y-intercept is the y value where the graph of the function crosses the y-axis. In other words, it's the y value when x = 0. To find the y-intercept, simply substitute 0 for x:

[tex]y = 3(0) + 75 \\ y = 75[/tex]

We find that the y-intercept is y = 75.

In the linear function y=3x+75, the slope 3 represents the rate of cost per mile to tow a car ($3/mile), and the y-intercept 75 is the basic fee for the towing service. So, the total cost is a fixed fee of $75 plus $3 per mile towed.

In the linear function y = 3x + 75, the slope, represented by '3', is the cost in dollars per mile to tow a car. This means for each mile the car is towed, the cost increases by $3.

The y-intercept, represented by '75', is the basic fee for the service, or the cost in dollars that one must pay even if the car isn't towed any miles.

So, to interpret the equation function: the total cost (y) of having a car towed is composed of a fixed fee of $75 plus an additional $3 for each mile (x) the car is towed.

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Answer:

[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]  

[tex]p_v =P(Z>0.8104)=0.209[/tex]  

If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{1}=23[/tex] represent the number of women favored the increase

[tex]X_{2}=19[/tex] represent the number of men favored the increase

[tex]n_{1}=50[/tex] sample 1 selected

[tex]n_{2}=50[/tex] sample 2 selected

[tex]p_{1}=\frac{23}{50}=0.46[/tex] represent the proportion of women favored the increase

[tex]p_{2}=\frac{19}{50}=0.38[/tex] represent the proportion of men favored the increase

z would represent the statistic (variable of interest)  

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

[tex]\alpha=0.1[/tex] represent the significance level

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if that a larger proportion of women favor the increase than men, the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} \leq p_{2}[/tex]  

Alternative hypothesis:[tex]p_{1} > p_{2}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{23+19}{50+50}=0.42[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]  

4) Statistical decision

We can calculate the p value for this test.  

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

[tex]p_v =P(Z>0.8104)=0.209[/tex]  

If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.  

Answer:

a) Null hypothesis:[tex]\mu \geq 40[/tex]  

Alternative hypothesis:[tex]\mu < 40[/tex]  

b) The rejection zone for this test would be:

[tex] (-\infty, -2.36)[/tex]

c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=39.2[/tex] represent the sample mean  

[tex]s=8[/tex] represent the sample standard deviation

[tex]n=100[/tex] sample size  

[tex]\mu_o =40[/tex] represent the value that we want to test

[tex]\alpha=0.01[/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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 40, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 40[/tex]  

Alternative hypothesis:[tex]\mu < 40[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{39.2-40}{\frac{8}{\sqrt{100}}}=-1[/tex]    

Part b: Rejection zone

On this case we are conducting a left tailed test so we need to find first the degrees of freedom for the distribution given by:

[tex]df=n-1=100-1=99[/tex]

Now we need to find a critical value on th t distribution with 99 degrees of freedom that accumulates 0.01 of the area on the right tail and this value is given by [tex]t_{crit}=-2.36[/tex]

And we can find it using the following excel code: "=T.INV(0.01,99)"

So then the rejection zone for this test would be:

[tex] (-\infty, -2.36)[/tex]

Part c: P-value and conclusion

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

[tex]p_v =P(t_{(99)}<-1)=0.160[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.  

Answer:

We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 minutes

Sample mean, [tex]\bar{x}[/tex] = 31.5 minutes

Sample size, n = 41

Alpha, α = 0.05

Sample standard deviation, s = 3.5 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{31.5 - 30}{\frac{3.5}{\sqrt{41}} } = 2.744[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

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

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Answer:

The equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Step-by-step explanation:

Given points are (-20,5) and (-36, 9)

Now to find the equation of the line passes through these points

Let [tex](x_{1},y_{1})[/tex] and  [tex](x_{2},y_{2})[/tex] be the two given points  (-20,5) and (-36, 9) respectively.

To find slope

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{9-5}{-36-(-20)}[/tex]

[tex]m=\frac{4}{-36+20}[/tex]

[tex]m=\frac{4}{-16}[/tex]

[tex]m=-\frac{1}{4}[/tex]

Therefore  [tex]m=-\frac{1}{4}[/tex]

The equation of the line is of thr form y=mx+c

The point (-20,5) passes through the above line and  [tex]m=-\frac{1}{4}[/tex]

[tex]5=-\frac{1}{4}(-20)+c[/tex]

[tex]5=5+c[/tex]

[tex]c=0[/tex]

[tex]y=-\frac{1}{4}x+0[/tex]

Therefore  [tex]y=-\frac{1}{4}x[/tex]

Therefore the equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Answer:

Part (a): 25 people became ill with the flu when the epidemic​ began.

Part (b): 1373 people were ill by the end of the fourth​ week.

Part (c): The limiting size of the population that becomes​ ill is 107,000.

Step-by-step explanation:

Consider the provided logistic growth function

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-t}}[/tex]

Part (a) How many people became ill with the flu when the epidemic​ began?

Substitute t = 0 in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-0}}[/tex]

[tex]f (t )=\frac{ 107,000}{4201}[/tex]

[tex]f (t )=25.47[/tex]

Hence, 25 people became ill with the flu when the epidemic​ began.

Part (b) How many people were ill by the end of the fourth​ week?

Substitute t = 4 in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-4}}[/tex]

[tex]f (t )=1373.10[/tex]

Hence, 1373 people were ill by the end of the fourth​ week.

Part (c) What is the limiting size of the population that becomes​ ill?

Substitute t = ∞ in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-\infty}}[/tex]

[tex]f (t )=\frac{ 107,000}{1 +0}[/tex]

[tex]f (t )=107,000[/tex]

Hence, the limiting size of the population that becomes​ ill is 107,000.

Using the logistic growth function, the number of sick people at the start of the epidemic and at the end of the fourth week can be found by substituting the appropriate values for time into the function. The limiting size of the potentially ill population is the maximum value of the function, which is 107,000.

The logistic growth function provided describes the number of people, represented by f(t), who have become ill with influenza, t weeks after its initial outbreak. The function is given as f(t) = 107000/(1 + 4200 * e-t).

a. At the beginning of the epidemic (i.e., when t=0), the number of ill people, f(0), can be calculated by substituting 0 for t in the equation, resulting in 107,000/(1+4200) people.

b. By the end of the fourth week (i.e., when t=4), the number of ill people, f(4), can be calculated by substituting 4 for t in the equation.

c. The limiting size of the population that becomes ill is represented by the maximum value the growth function can take, which in this case is 107,000 (as seen by the numerator of the function). This means that illness will eventually spread to reach a potential maximum of 107,000 people in this specific community.

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Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Number of colors that the arrow can land on = 8

We need to find the degrees of freedom for a goodness of fit test of the fairness of this spinner.

As we know that degree of freedom is 1 less than the size of sample.

So, Mathematically, it is expressed as below:

So, Degree of freedom would be

[tex]v=n-1\\\\v=8-1\\\\v=7[/tex]

Hence, Option 'c' is correct.

Answer:

[tex]\frac{dV}{dt}=525 cm^{3}/min[/tex]

Step-by-step explanation:

The volume of a sphere is:

[tex]V=\frac{4}{3}\pi r^{3}[/tex] (1)

We know that:

If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.

[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex] (2)

We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.

Solving (1) for r, we have:

[tex]r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm[/tex]

Finally dV/dt will be:

[tex]\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min[/tex]

I hope it helps you!

Answer:

The correct option is B) 0.0228

Step-by-step explanation:

Consider the provided information.

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

Thus, the value of n = 100, [tex]\bar x=3.1[/tex] and σ = 0.5

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

The null and alternative hypotheses are,

[tex]H_0:\mu\leq3[/tex] and [tex]H_a:\mu>3[/tex]

Compute the test statistic [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}[/tex]

[tex]z=\frac{0.1}{0.05}[/tex]

[tex]z=2[/tex]

By using the table.

P value = P(Z>2) = 0.0228

Hence, the correct option is B) 0.0228

Answer:

Step-by-step explanation:

To answer this question, First, we define a direction in space, which defines

the North Pole of each planet.

We also assume that the direction is not at 90° perpendicular to the axis of any two planet.

Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.

If we refer to the direction of the planet in the diagram, we discover that,

1, A ≥ A

2, If A ≥ B and B≥ A,

Then A = B

3, If A ≥ B, and B ≥ C , then A ≥ C

4, Either A ≥ B or B ≥ A

It should also be noted that the array of order above has a unique maximal element  (M). This is the only  planet whose North Pole is not visible from another.

If we now consider a sphere of the same radius as the planets. Remove

from it all North Poles defined by directions that are perpendicular

to the axis of two of the planets. This is a set of area zero.

For every other point on this sphere, there exists a direction in

space that makes it the North Pole, and for that direction, there

exists a unique North Pole on one of the planets which is not visible

from the others. Hence the total area of invisible points is equal

to the area of this sphere, which in turn is the area of one of the  planets.

Answer:

92.79% probability that the mean clock life would be greater than 15.1 years.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/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.

In this problem, we have that:

[tex]\mu = 16, \sigma = 4, n = 42, s = \frac{4}{\sqrt{42}} = 0.6172[/tex]

What is the probability that the mean clock life would be greater than 15.1 years?

This probability is 1 subtracted by the pvalue of Z when [tex]X = 15.1[/tex]. So:

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

[tex]Z = \frac{15.1 - 16}{0.6172}[/tex]

[tex]Z = -1.46[/tex]

[tex]Z = -1.46[/tex] has a pvalue of 0.0721.

So there is a 1-0.0721 = 0.9279 = 92.79% probability that the mean clock life would be greater than 15.1 years.

Final answer:

Using the Central Limit Theorem and the given standard deviation and mean, the standard error was calculated, and then the z-score was determined. After finding the z-score, the area to the right of it on the standard normal distribution curve gives the probability that the mean clock life is greater than 15.1 years, which is 0.9274.

Explanation:

To find the probability that the mean clock life of a sample of 42 wall clocks is greater than 15.1 years, we can use the Central Limit Theorem which tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we are dealing with a sample of 42, which is generally considered sufficiently large, the sampling distribution of the sample mean is normally distributed. Given the population mean (μ) is 16 years, the standard deviation (σ) is 4 years, the sample size (n) is 42, and we want to find the probability of the sample mean (μ_x) > 15.1 years, we first need to find the standard error (SE) which is σ/sqrt(n). Then we can find the z-score which is (15.1 - μ) / SE. Lastly, we use the z-score to find the probability from z-tables or a calculator with normal distribution functions.

The standard error (SE) is:
SE = 4 / sqrt(42) ≈ 0.6172
The z-score (z) is:
z = (15.1 - 16) / 0.6172 ≈ -1.4573

To find the probability of the sample mean being greater than 15.1 years, we want the area to the right of the z-score on the standard normal distribution curve. Looking up the z-score of -1.4573 on the z-table or using a normal distribution calculator gives us the probability to the left of the z-score. We subtract this value from 1 to find the probability to the right:
P(X > 15.1) = 1 - P(Z < -1.4573) ≈ 1 - 0.0726 = 0.9274

Therefore, the probability that the mean clock life is greater than 15.1 years is 0.9274 when rounded to four decimal places.

Answer:

We conclude that the actual average cost per workbook is higher than $27.50.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $27.50

Sample mean, [tex]\bar{x}[/tex] = $28.90

Sample size, n = 44

Alpha, α = 0.05

Population standard deviation, σ = $5.00

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 27.50\text{ dollars}\\H_A: \mu > 27.50\text{ dollars}[/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{28.90 - 27.50}{\frac{5.00}{\sqrt{44}} } = 1.8573[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

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

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the actual average cost per workbook is higher than $27.50.

Answer:

1a) Plot on the graph.

1b) Its proportional if y (weight) varies directly with variation in x(age) and fits the form y=k*x

Step-by-step explanation:

1a)

draw graph with x axis in increments of 1 month

draw y-axis in increments of 1 pound.

put 1 dot at (1, 4)

put 1 dot at (2,7)

connect the 2 dots, and continue the line to hit one of the axes.

It will not hit the origin.

1b) show that y=kx

4=k*1 , so k=4

7=k*2 , so k=3.5

Since we have 2 different values for k (4 & 3.5), it's not proportional.

Answer:

The equation of the line with slope is 3  and point (2,17) is [tex]y=3x+11[/tex]

Step-by-step explanation:

Given slope is 3 ie., m=3 and point (2,17)

To find the equation of the line with slope and a point:

Using slope point formula

[tex]y-y_{1}=m(x-x_{1})[/tex]

Let the point (2,17) be [tex](x_{1},y_{1})[/tex]

Substitute the values of m=3 and point (2,17)

[tex]y-y_{1}=m(x-x_{1})[/tex]

[tex]y-17=3(x-2)[/tex]

[tex]y-17=3x-6[/tex]

[tex]y=3x-6+17[/tex]

[tex]y=3x+11[/tex]

Therefore the equation of the line is [tex]y=3x+11[/tex]

Answer:

a) Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

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

[tex]s[/tex] represent the sample deviation

[tex]n=18[/tex] sample size  

[tex]\mu_o =10[/tex] represent the value that we want to test

[tex]\alpha[/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 true mean is at least 10 hours, the system of hypothesis would be:  

Part a

Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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)  

Part b

For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Part c

For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

Part d

For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.

This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.

The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:

The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.

With t = -23 and α = 0.05,  this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.

Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.

Finally,  with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.

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A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

A. 0.037773 to 0.202227

B. 3.7773% to 20.2227%

C. 59.98786% to 61.01214%

D. 51.7773% to 68.2227%

E. 10.988% to 13.012%

Answer: E. 10.988% to 13.012%

Step-by-step explanation:

Given;

Mean x= 12%

Standard deviation r = 4%

Number of samples tested n = 60

Confidence interval is 95%

Z' = t(0.025)= 1.96

Confidence interval = x +/- Z'(r/√n)

= 12% +/- 1.96(4%/√60)

= 12% +/- 0.01214%

Confidence interval= (10.988% to 13.012%)

Answer:

Step-by-step explanation:

Given

mean [tex]\mu =20\ ounces[/tex]

standard deviation [tex]\sigma =1\ ounce[/tex]

The no of sample boxes weigh Every morning is 25

Average weight is 1 % more than average

i.e. [tex]20+20\times 0.1=20.2 ounce[/tex]

The company re-calibrates the machine

[tex]P\left ( \bar{x}> 20.2\right )=P\left ( z-\frac{20.2-20}{\frac{1}{\sqrt{25}}}\right )[/tex]

[tex]=P\left ( z> 1\right )=1-P\left ( z\leq 1\right )[/tex]

[tex]=1-0.8413[/tex]

[tex]=0.1587[/tex]

Therefore the Probability that the average weight of box is more than 20.2 ounce is 0.1587

No of days the machine is expected to re-calibrate is [tex]100\times 0.1587=16\ days[/tex]

Answer:

True

Step-by-step explanation:

In Mathematics, A radical symbol is defined as √

It is an expression that is uses a root, such as the square root (√ ) or cube root ( ∛ )

To fully show a radical expression:

[tex]\sqrt{A}[/tex]

A is the radicand

√ is the radical symbol

Here, the degree is 2.

Quotient Rule for Radicals

For non- negative real numbers, x and y:

[tex]\frac{\sqrt[n]{x}}{\sqrt[n]{y} }[/tex] = [tex]\sqrt[n]{\frac{x}{y}}[/tex]

Example :

[tex]\sqrt{\frac{36}{9}}[/tex]= [tex]\frac{\sqrt{36}}{\sqrt{9} }[/tex] = [tex]\frac{6}{3}[/tex] = 2

So, in simplifying radicals, quotient rule is used as demonstrated above. So the condition is true