A marketing consultant was hired to visit a random sample of five sporting goods stores across the state of California. Each store was part of a large franchise of sporting goods stores. The consultant taught the managers of each store better ways to advertise and display their goods.
The net sales for 1 month before and 1 month after the consultant's visit were recorded as follows for each store (in thousands of dollars):

Before visit: 57.1 94.6 49.2 77.4 43.2
After visit: 63.5 101.8 57.8 81.2 41.9

Do the data indicate that the average net sales improved? (Use a= 0.05)

Answer: don’t use k for the other numbers

Step-by-step explanation:

Answer:

6k -5

Step-by-step explanation:

19-6(-k+4)

Distribute

19 -6*-k -6*4

19 +6k -24

Combine like terms

6k +19-24

6k -5

Answer:

1. y = f(8)

2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

Step-by-step explanation:

The concentration of the drug in the bloodstream at t hours is:

y = f(t)

1. What is the concentration of the drug in the bloodstream at t= 8 hours?

At t hours, y = f(t)

So at 8 hours, y = f(8)

2. During what time interval is the drug level in the bloodstream increasing? Decreasing?

A function f(t) is increasing when

f'(t) > 0

And is decreasing when

f'(t) < 0

So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

(1) The concentration  of drug in the bloodstream in 8 hours is given by

[tex]\rm \bold{y = f (8)}[/tex]

(2) The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y' = f'(t) <0[/tex] the drug level of the bloodstream is decreasing.

When a certain prescription drug is taken orally by an adult.

the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y=f(t)

To be determined

(1) The concentration of the drug in the bloodstream at t= 8 hours

(2) During what time interval is the drug level in the bloodstream increasing or deceasing

The amount of the drug (in mg/L) in the bloodstream at t hours is given by the function

y=f(t).......(1)

(1)  The concentration  of drug in the bloodstream in 8 hours is given by putting t= 8 in the equation (1) which can be formulated as below

[tex]\rm y = f (8)[/tex]

(2) From the definition of increasing and decreasing function we can write that

[tex]\rm y = f(x) \; is \; increasing \; when \; f' (x)>0 \\and\; y = f(x) ; is \; decreasing \; when \; f' (x)<0 \\\\\\\\[/tex]

By the definition of increasing and decreasing function we can say that

The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y'=f'(t) <0[/tex] the drug level of the blood stream is decreasing.

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Answer: I had a stroke trying to read and understand this sorry

Step-by-step explanation:

Answer:

$60.

Step-by-step explanation:

x / 2 - 10 = 0

x / 2 = 10

x = 20

Before she spent her money:

x / 2 - 10 = 20

x / 2 = 30

x = 60

She had $60 before she went shopping.

Feel free to let me know if you need more help. :)

Answer:100 people

Step-by-step explanation:

140 people x 5 ladle/1 person = 700 ladles full

700 ladles x 1 person/7 ladles = 100 people fed

Answer:

5

Step-by-step explanation:

It be 5 because its in the middle

3,4,7,8

Step-by-step explanation:

1 2 3 4 5 6 7 8 9

3 4 7 8

Answer:

C) 192.11

Step-by-step explanation:

To find the mean of any group of numbers, you simply add up all the numbers and then divide by how many numbers there are.

Using a calculator or a pencil, add the following:

180 + 220 + 240 + 196 + 175 + 183 + 195 + 140 + 200

= 1,729

The last step is to divide 1,729 by 9:

1,729 ÷ 9 = 192.11

Answer:

Find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Step-by-step explanation:

Checking for Normality means basically checking if one's data distribution approximates a normal distribution.

A normal distribution is represented by a bell-shaped curve, peaking around the mean, indicating that all of the data spreads out from the mean.

Th original aim of the teacher is to check the effects of the particular added unit on the performance of students in the subject.

The teacher goes about this by testing the students before and after learning the unit.

The best way to compare of course, is to take a difference of the test scores for different samples. This first gives the idea of whether the newly introduced unit affects performance.

This set of differences is then checked for normality.

So, the best manner to make a plot of these differences. Like we mentioned earlier, a normal distribution is bell shaped. So, the plot of these differences would be a bell shaped curve if the distribution was normal and we wouldn't get a bell shaped curve if the distribution wasn't normal.

Checking for outliers help to eliminate part of data that can totally scatter the regular behaviour of the data distribution.

So, the best way for the teacher to check for normality is to find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Hope this Helps!!!

Since the data include the pretest and post test scores of each student or subject, then, it is approached as a paired t test. Hence, checking for normality would require obtaining the difference between the two scores for each subject, then make a plot of the differences and check for outliers.

Hence, the most appropriate check for normality in this scenario is to check for outliers in the plot made from the score difference.

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I suppose you're supposed to prove that set intersection is distributive across a union,

[tex]A\cap(B\cup C)=(A\cap B)\cup(A\cap C)[/tex]

Two sets are equal if they are subsets of one another. To prove a set [tex]X[/tex] is a subset of another set [tex]Y[/tex], you have to show that any element [tex]x\in X[/tex] also belongs to [tex]Y[/tex].

Let [tex]x\in A\cap(B\cup C)[/tex]. By definition of intersection, both [tex]x\in A[/tex] and [tex]x\in B\cup C[/tex]. By definition of union, either [tex]x\in B[/tex] or [tex]x\in C[/tex]. If [tex]x\in B[/tex], then clearly [tex]x\in A\cap B[/tex]; if [tex]x\in C[/tex], then [tex]x\in A\cap C[/tex]. Either way, [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Hence [tex]A\cap(B\cup C)\subseteq(A\cap B)\cup(B\cap C)[/tex].

The proof in the other direction uses the same sort of reasoning. Let [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Then either [tex]x\in A\cap B[/tex] or [tex]x\in A\cap C[/tex]. If [tex]x\in A\cap B[/tex], then both [tex]x\in A[/tex] and [tex]x\in B[/tex]; if [tex]x\in A\cap C[/tex], then both [tex]x\in A[/tex] and [tex]x\in C[/tex]. So certainly [tex]x\in A[/tex], and either [tex]x\in B[/tex] or [tex]x\in C[/tex] so that [tex]x\in B\cup C[/tex]. Hence [tex](A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C)[/tex].

Both sets are subsets of one another, so they are equal.

Answer:

b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Step-by-step explanation:

The confidence interval is an estimation for the true population parameter, calculated from the information of a sample of this population.

The parameter of the population will be within this interval with a certain degree of confidence.

a. There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Incorrect. The confidence interval gives only the probability that the true proportion (or population proportion) is within 17.4% and 24.8%, not the proportion of individual samples.

b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Correct.

c. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Incorrect. The confidence interval does not give information about another samples.

d. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year.

Incorrect. The confidence interval does not give information about another samples or sampling distributions.

Answer: £569.40p

Step-by-step explanation:

410÷£0.72=£569.40p

Answer:

£295.20

Step-by-step explanation:

€410 x 0.72 = £295.20

Answer:

  (1000 m)/(1 km) and (60 min)/(1 h)

Step-by-step explanation:

Put the unit you don't want in a position to cancel the unit in the given number. Write the fraction so that the equivalent amount of the unit you do want is on the other side of the fraction bar.

Here we have km/min with km in the numerator. To cancel that, we need a fraction with km in the denominator. We want meters (m) in the numerator, so we need a fraction that has a number of meters equivalent to 1 km. That will be ...

  (1000 m)/(1 km)

This is one of the conversion factors we will need to multiply by.

__

We also have "min" in the denominator. To cancel that, we need a conversion factor with min in the numerator. The unit we want in the denominator is h (hours), so we need an equivalent for hours and minutes. That would be 60 min = 1 h, so we write the conversion factor as ...

  (60 min)/(1 h)

So, our conversion factors are ...

  (1000 m)/(1 km) and (60 min)/(1 h)

_____

The converted number is ...

  (4 km/min)(1000 m/km)(60 min/h) = 240,000 m/h

__

Comment on conversion factors

As you can see, we write the fraction so equal amounts are in numerator and denominator. Since the amounts are equal, the value of the fraction is 1. Multiplying by 1 in this form doesn't change the original value, it only changes the units.

Answer:

6.67% probability that the result is a multiple of 5 and a multiple of 2

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Multiples of 2 AND 5

Between 1 and 15, the multiples of 2 are: 2,4,6,8,10,12,14

Between 1 and 15, the multiples of 5 are: 5,10,15

So only 10 is a multiply of both 2 and 5, so only one desired outcome, which means that [tex]D = 1[/tex]

Total outcomes:

Any number between 1 and 15, there are 15, so [tex]T = 15[/tex]

Probability:

[tex]p = \frac{D}{T} = \frac{1}{15} = 0.0667[/tex]

6.67% probability that the result is a multiple of 5 and a multiple of 2

Answer:

Step-by-step explanation:

3/5 or 9/15

Answer:

The value of the sample mean is 78.

Step-by-step explanation:

We are given that a sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8.

Also, the sample mean corresponds to a z score of 2.00.

Let [tex]\bar X[/tex] = sample mean

The z-score probability distribution for a sample mean is given by;

              Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 70

            [tex]\sigma[/tex] = standard deviation = 8

            n = sample size = 4

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, we are given that the sample mean corresponds to a z score of 2.00 for which we have to find the value of sample mean;

So, z-score formula is given by ;

                  z-score = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] = [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]

                      2.00 =  [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]

                      2.00 =  [tex]\frac{\bar X-70}{4 } }[/tex]

                     [tex]\bar X = 70+(2 \times 4)[/tex]

                     [tex]\bar X[/tex] = 70 + 8 = 78

Therefore, the value of the sample mean is 78.

The value of the sample mean is 78.

To find the value of the sample mean, we can use the formula:

sample mean = population mean + (z score * (standard deviation / square root of sample size))

In this case, the population mean is 70, the z score is 2.00, and the standard deviation is 8. Since the sample size is 4, we calculate the square root of 4, which is 2. Plugging these values into the formula gives us:

sample mean = 70 + (2.00 * (8 / 2)) = 70 + (2.00 * 4) = 70 + 8 = 78...

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c^2 = a^2 + b^2 - 2(ab)(cos C)

c^2 + 2(ab)(cos C) = a^2 + b^2

2(ab)(cos C) = a^2 + b^2 - c^2

cos C = (a^2 + b^2 - c^2) / 2ab - Answer choice E

Hope this helps! :)

To make COS the subject of a formula, the equation is rearranged such that COS is isolated. An example can be in the equation a = b cos(x) which can be rearranged as cos(x) = a/b. A complete formula is necessary for an accurate step-by-step guide.

To make COS the subject of a formula, it typically involves other known quantities represented by variables and constants. For example, in the equation a = b cos(x), we can make cos(x) the subject of the equation by rearranging it to: cos(x) = a/b. However, to provide a more accurate step-by-step guide, the complete formula is necessary. This principle can be applied to various trigonometric formulas so that COS becomes the main focus.

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Answer:

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

Yes, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

Suppose that we have:

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

For this case, we're provided that:

We want to determine if true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​% = 0.29 (converted percent to decimal).

Hypotheses:

Thus, the test is left tailed test.

where [tex]p_0[/tex] = 29% = 0.29 is the hypothesized mean value of population proportion.

The test statistic is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\ Z = \dfrac{715/2684 - 0.29}{\sqrt{\dfrac{0.29(1-0.29)}{2684}}} \approx -2.69[/tex]

The critical value of Z at level of significance 0.05 is -1.6449

Since the test statistic = -2.69 < critical value = -1.6449, so the test statistic lies in the rejection region (the rejection region for the left tailed test is all the values below critical value).

Thus, we reject the null hypothesis and accept the alternative hypothesis that the true population mean is less than 0.29.

Thus, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

The z-test statistic came out to be -2.69

Learn more about one-sample z-test for population proportion mean here:

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Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion teenage drivers = [tex]\frac{120}{576}[/tex] = 0.21

           n = sample of accidents = 576

           p = population percentage of all car accidents

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] ) = 0.95

P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] ) = 0.95

95% confidence interval for p = [[tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]]

  = [ [tex]0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex] , [tex]0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex] ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

Let p = percentage of teenagers who where involved in car accidents

So, Null Hypothesis, [tex]H_0[/tex] : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 18%    {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is One-sample z proportion statistics;

                              T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion teenage drivers = [tex]\frac{120}{576}[/tex] = 0.21

           n = sample of accidents = 576

So, test statistics  =  [tex]\frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}[/tex]  

                              =  1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, 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 the percentage of teenagers has changed since you join the company.

Answer: B The width of the interval would increase

Step-by-step explanation:

i just took this and go the answers back

Decreasing the sample size from 100 to 50 would widen the 95 percent confidence interval for the difference in mean seat widths between two airlines. This is because smaller samples have more variability, requiring a larger interval to capture the population mean with the same level of certainty.

If the consumer group used a sample size of 50 instead of 100 for each airline, that would increase the width of the 95 percent confidence interval for the difference in population mean widths of seats between the two airlines. This is because smaller sample sizes result in more variability, requiring a wider interval to capture the true population mean with the same level of certainty.

As with the unoccupied seats example, where the sample mean of 11.6 and standard deviation of 4.1 were used to form a confidence interval, the size of the interval is dependent on the variability within the sample, and smaller samples generally have higher variability. Similarly in the case of exam scores, with a lower confidence level of 90 percent, a narrower interval is needed compared to a higher confidence level of 95 percent.

Therefore, in effect, decreasing the sample size from 100 to 50 would make the confidence interval wider, as more variability is expected and a larger interval is needed to capture the true population mean with a 95 percent confidence level.

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Answer:

[tex]s^2=1.32[/tex]

Step-by-step explanation:

-This is a binomial probability distribution problem.

-Given that p=0.86 and n=11, the sample variance can be calculated using the formula:

[tex]\sigma^2=np(1-p)[/tex]

#We substitute the given parameters in the formula to solve for variance:

[tex]\sigma^2=np(1-p)\\\\\\=11\times 0.86(1-0.86)\\\\=1.324\approx1.32[/tex]

Hence, the sample variance is 1.32

Final answer:

The variance in a sample of 11 flights that have arrived on time at Denver International Airport, where 86% of flights are on time, is calculated using a binomial distribution with the formula variance = np(1-p), resulting in a variance of 1.33 after rounding to two decimal places.

Explanation:

The subject of this question is about finding the variance in a sample of flights that have arrived on time at Denver International Airport, given that 86% of recent flights have arrived on time and that a sample of 11 flights is studied. To calculate the variance for a binomial distribution, which is applicable in this context because each flight can either be on time or not, we use the formula:

Variance = np(1-p).

Where 'n' is the number of trials (or flights, in this case), which is 11, and 'p' is the probability of success on each trial (a flight arriving on time), which is 0.86. Thus, the variance for the 11 flights can be calculated as follows:

Variance = 11 × 0.86 × (1 - 0.86) = 11 × 0.86 × 0.14 = 1.3284.

After rounding to two decimal places, the variance for a sample of 11 flights is 1.33.

Answer:

g(-1) = -3

Step-by-step explanation:

g(-1) is the y value of the graph when x = -1

when x=-1 y=-3

g(-1) = -3

Answer:

Cube surface area = 6 * side^2

150 square inches = 6 * side^2

sq root (side) = 25

side = 5 inches

Step-by-step explanation:

Step-by-step explanation:

(2575÷25)(1000÷25) = 10340 when reduced to the simplest form. As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 25751000 is also equal to 22340 when expressed as a mixed number.

Answer:

$371.39

Step-by-step explanation:

150 * .12+150 = 168

168 * .12+168 = 188.16

188.16 * .12+188.16 = 210.7392

210.7392 * .12+210.7392 = 236.027904

264.3512525 * .12+264.3512525 = 296.0734028

296.0734028 * .12+296.0734028 = 331.6022111

331.6022111 * .12+331.6022111 = 371.3944764

you multiply your current number by 12% and add that to the number, the last number i rounded for the answer as you can see

Answer:

(a) P (Z < 2.36) = 0.9909                    (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112                      (d) P (1.13 < Z > 3.35)  = 0.1288

(e) P (-0.77< Z > -0.55)  = 0.0705       (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995                   (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, [tex]X \sim N (\mu, \sigma^{2})[/tex], then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, [tex]Z \sim N (0, 1)[/tex].

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.  The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

                   ≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

                   = 1 - 0.99086

                   = 0.00914

                   ≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

                   ≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

                            = 0.99960 - 0.87076

                            = 0.12884

                            ≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35)  is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

                                = 0.29116 - 0.22065

                                = 0.07051

                                ≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55)  is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

             = 1 - 0.99865

             = 0.00135

             ≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

                    = 0.99948

                    ≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

                   ≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

To find the probabilities, we use the standard normal distribution table or a calculator to calculate probabilities for a standard normal distribution. We calculate each probability step by step and round the answers to four decimal places.

To determine the probabilities, we will use the standard normal distribution table or a calculator that can calculate probabilities for a standard normal distribution.

(a) P(z < 2.36) = 0.9900

(b) P(z > 2.36) = 1 - P(z < 2.36) = 1 - 0.9900 = 0.0100

(c) P(z < -1.22) = 0.1103

(d) P(1.13 < z < 3.35) = P(z < 3.35) - P(z < 1.13) = 0.9993 - 0.8708 = 0.1285

(e) P(-0.77 < z < -0.55) = P(z < -0.55) - P(z < -0.77) = 0.2896 - 0.2823 = 0.0073

(f) P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013

(g) P(z > -3.28) = 1 - P(z < -3.28) = 1 - 0.0005 = 0.9995

(h) P(z < 4.98) = 1 - P(z > 4.98) = 1 - 0.0000 = 1.0000

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Answer:

[tex]x+3y=-27[/tex]

Step-by-step explanation:

We are given that

[tex]y=(-\frac{1}{3})x-9[/tex]

We have to find the  standard form of given equation

[tex]y=\frac{-x-27}{3}[/tex]

[tex]3y=-x-27[/tex]

By using multiplication property of equality

[tex]x+3y=-27[/tex]

We know that

Standard form of equation

[tex]ax+by=c[/tex]

Therefore, the standard form of given equation  is given by

[tex]x+3y=-27[/tex]

The equation y = -1/3x - 9 can be converted to standard form by eliminating fractions and rearranging. The final equation in standard form is x - 3y = 27.

The equation you provided is already in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. However, you're asked to convert this equation into standard form, which is Ax + By = C, with A, B, and C being integers, and A and B not both equal to zero.

To convert the given equation y = -1/3x - 9 into standard form, we first need to eliminate the fractions. We can achieve this by multiplying every term by -3, giving us 3y = x + 27. To make it fit the standard form, we can rearrange as -x + 3y = -27.

Remember, standard form shouldn't have any negatives in front of the x term, so we multiply everything by -1. The final equation in standard form is x - 3y = 27.

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Answer:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Step-by-step explanation:

Notation

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

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

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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)

We can calculate the mean and the sample deviation we can use the following formulas:  

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

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=15.55[/tex]

The sample deviation calculated [tex]s=0.278[/tex]

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=8-1=7[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that [tex]t_{\alpha/2}=2.997[/tex]

And the confidence interval is given by:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Answer:

Step-by-step explanation:

(a) A parameter of a population measures the characteristics of the population, In the question the proportion of all the persons who have health insurance and the mean of the entire dollar amount that Americans spent on health care in the past year measure the population.

Invariably, the proportion of persons having health insurance, and the mean dollars spent on health care for all Americans are the population parameter

(b)  A statistics  measure the characteristics of the sample.

In the question, the sample of 1500 Americans are considered to estimate the proportion of all Americans, proportion of all the persons who have health insurance among 1500 and the sample mean of all the dollar amounts that the selected Americans spent on healthy care in the past year describe  the sample.

Invariably, the sample proportion of persons having health insurance, and the mean dollars spent on health care for 1500 selected Americans are sample statistics

Answer: Top Right, A rectangle has all the properties of a square.

Step-by-step explanation: Nobody how hard you try to make a rectangle a square, it won't work. A rectangle cannot have 4 equal slides.

Given:

Given that Brinson's family is going camping. Their tent is shaped like a rectangular pyramid.

The volume of the tent is 6000 cubic inches.

The area of the base of the tent is 1200 square inches.

We need to determine the height of the tent.

Height of the tent:

The height of the tent can be determined using the formula,

[tex]V=\frac{1}{3}Bh[/tex]

where B is the area of the base and h is the height of the pyramid.

Substituting V = 6000 and B = 1200, we get;

[tex]6000=\frac{1}{3}(1200)h[/tex]

[tex]6000=400h[/tex]

   [tex]15=h[/tex]

Thus, the height of the tent is 15 inches.

Answer:

  yes

Step-by-step explanation:

The 90-th percentile of a normal distribution corresponds to a z-score of 1.282. Bob's z-score is above that, so he is definitely in the top 10%.

_____

If you haven't memorized the percentiles associated with the normal distribution, it is convenient to use a calculator or table of values.

In this case, ob is correct in saying that his score is in the top 10 percent.

To determine if Bob's z-score corresponds to a score in the top 10 percent of the distribution, we need to find the percentile associated with his z-score.

We can then compare this percentile to 90%, as the top 10% corresponds to the highest scores.

Using a standard normal distribution table or a calculator, we find that a z-score of 1.52 corresponds approximately to the 93rd percentile.

This means that 93% of the scores are below Bob's score, indicating that he is indeed in the top 10% of the class.

So, Bob is correct in saying that his score is in the top 10 percent.