Problem 3: Sampling methods
Willy wants to find what percent of students at his school drink the milk after
they finish their cereal. He is considering the following sampling methods:
QUESTION
He selects every tenth person who enters the school. What type of sampling
is this?
Choose 1 answer:

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:

12.36

Step-by-step explanation:

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).

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

We want to find a recursive formula for f(n) = 64 + 6n

The recursive formula is f(n) = f(n - 1) + 6

First, a recursive formula is a formula that gives the value of f(n) in relation to the value of f(n - 1) or other previous terms on the sequence.

We know that:

f(n) = 64 + 6n

f(n - 1) = 64 + 6*(n - 1) = 64 + 6*n - 6

Then we can rewrite:

f(n) = 64 + 6n + 6 - 6

f(n) = ( 64 + 6n - 6) + 6

And the thing inside the parenthesis is equal to f(n - 1)

Then we have:

f(n) = f(n - 1) + 6

This is the recursive formula we wanted to get.

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Answer:

Check the explanation

Step-by-step explanation:

Number of transactions in a day is sum of number of withdrawals and number of deposits. So,

Number of transactions in a day, Z = X + Y

Moment Generating function of Z is,

T+1

Expected number of transactions in a day = E[Z]

[tex]= \frac{\mathrm{d} }{\mathrm{d} t}_{t=0}M_Z(t) = (r+1)\left ( \frac{p}{1-qe^t} \right )^{r} * \frac{-p}{(1-qe^t)^2} * (-qe^t) for t = 0[/tex]

[tex]= (r+1)\left ( \frac{p}{1-qe^0} \right )^{r} * \frac{-p}{(1-qe^0)^2} * (-qe^0)[/tex]

[tex]= (r+1)\left ( \frac{p}{1-q} \right )^{r} * \frac{-p}{(1-q)^2} * (-q)[/tex]

[tex]= (r+1)\left ( \frac{p}{p} \right )^{r} * \frac{-p}{(p)^2} * (-q)[/tex]

[tex]= \frac{(r+1)q}{p}[/tex]

Answer:

6 Years

Step-by-step explanation:

Orlando invests $1000 at 6% annual interest compounded daily.

Orlando's investment = [tex]A=1000(1+\frac{0.06}{365})^{(365\times t)}[/tex]

Bernadette invests $1000 at 7% simple interest.

Bernadette's investment = A = 1000(1+0.07×t)

By trail and error method we will use t = 5

Bernadette's investment will be after 5 years

1000(1 + 0.07 × 5)

= 1000(1 + 0.35)

= 1000 × 1.35

= $1350

Orlando's investment after 5 years

[tex]A=1000(1+\frac{0.06}{365})^{(365\times 5)}[/tex]

   = [tex]1000(1+0.000164)^{1825}[/tex]

  = [tex]1000(1.000164)^{1825}[/tex]

  = 1000(1.349826)

  = 1349.825527 ≈ $1349.83

After 5 years Orlando's investment will not be more than Bernadette's.

Therefore, when we use t = 6

After 6 years Orlando's investment will be = $1433.29

and Bernadette's investment will be = $1420

So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.

Ok, what's the full question here

Answer:

10 (3x+7)

Step-by-step explanation:

factor out 10

10 (3x+7)

Answer:

M(Mx, My), R(Rx, Ry), S(Sx, Sy)

M is midpoint of RS

=> Rx + Sx = 2Mx => Rx = 2Mx - Sx = 2 x 8 - 9 = 7

=> Ry + Sy = 2My => Ry = 2My - Sy = 2 x 7 - 5 = 9

=> R(7, 9)

Hope this helps!

:)

Answer:

(7,9)

Step-by-step explanation:

Let R:(x,y)

(x+9)/2, (y+5)/2 = 8 ,7

x + 9 = 16

x = 7

y + 5 = 14

y = 9

R: (7,9)

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:

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:

The answer to this question can be described as follows:

Step-by-step explanation:

In the outcome of Jackson isn't legitimate as product A and product B were n’t considered equal in watering the plants. It is the preparation for all two to be compared, Product A with only a region of the backyard, which receives direct energy, and Product B will be maintained regularly by the very same area of the yards, that is in minimal lighting.

No, The conclusion done by Jackson's is not valid for this situation.

Effective Soiling depth is the depth of soil material  which is above the layer that impedes movement of air and water and growth in plant roots.

According to outcome of Jackson that isn't legitimate as product A and product B were considered equal in watering the plants. Then it is the preparation of two to be compare with product B, Product A with only a region of the backyard, which receives direct energy, and Product B will  maintained regularly very same area of the yards, that is with minimal lighting.

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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.

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: I think c,e

Step-by-step explanation:Sorry if it wrong

Answer:

C

Step-by-step explanation:

IM DORA

Answer:

B

Step-by-step explanation:

For one-way ANOVA test results to be reliable, it has to be assumed that standard deviations of populations are equal. Here standard deviations of populatinos are not mentioned.

Answer:

x = 12

Step-by-step explanation:

20° + (6x - 2) = 90°

18° + 6x = 90°

6x = 90° - 18°

6x = 72°

x = 72°/6

x = 12

Answer:

Hope it helps buddy! Mark as brainlist please.

In response to the question, we may say that the height of the cylinder is 6.

A cylinder is a three-dimensiοnal geοmetric shape made up οf twο parallel cοngruent circular bases and a curving surface cοnnecting the twο bases. The bases οf a cylinder are always perpendicular tο its axis, which is an imaginary straight line passing thrοugh the centre οf bοth bases. The vοlume οf a cylinder is equal tο the prοduct οf its base area and height.

A cylinder's volume is computed as V = r²h,

where

"V" represents the volume,

"r" represents the radius of the base, and

"h" represents the height of the cylinder.

Thus,  

307.9 = 7²h

307.9 = 49h

h = 307.9/49

h 6.283

h = 6     (rounding to the nearest whole)

Thus, The height of the cylinder is 6.

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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:

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.

The most expensive type of life insurance is cash-value (whole) life insurance due to its integrated savings component, investment aspect, and the management involved. These features add complexity and cost to the policy, making it more expensive than term life insurance, which only offers a death benefit.

The most expensive type of life insurance is typically cash-value (whole) life insurance. This kind of policy not only provides a death benefit to beneficiaries upon the insured individual's death but also includes a savings component known as the cash value. Over time, the policy accumulates a cash value that the policyholder can borrow against or even withdraw. Part of the premium goes towards life insurance, while another part is invested by the company, often in a conservative mix of investments. This investment aspect and the management involved make whole life insurance more expensive compared to term life insurance which only provides a death benefit without a savings component.

It is also worth noting that individuals purchasing life insurance are likely to have more knowledge about their own health risks than the insurance company. This information asymmetry can impact the pricing of life insurance policies. Moreover, the fundamental law of insurance states that the average person's premium payments over time must cover the claims, the insurer's costs, and profits. Therefore, insurance premiums, particularly for policies with cash values, reflect the cost of investment management and the integrated savings aspects, which can contribute to their higher costs.

The easy answer: If, say, B is the zero matrix, i.e.

[tex]B=\begin{bmatrix}0&0\\0&0\end{bmatrix}[/tex]

then

[tex]2A=\begin{bmatrix}5&0\\-1&2\end{bmatrix}\implies A=\begin{bmatrix}\dfrac52&0\\\\-\dfrac12&1\end{bmatrix}[/tex]

Answer:

You should claim 10.5325 years on your warranty.

Step-by-step explanation:

Problems of normally distributed samples are 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 = 13, \sigma = 1.5[/tex]

Want's to replace no more than 5% of the products.

This means that the warranty should be 5th percentile, that is, the value of X when Z has a pvalue of 0.05. So X when Z = -1.645.

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

[tex]-1.645 = \frac{X - 13}{1.5}[/tex]

[tex]X - 13 = -1.645*1.5[/tex]

[tex]X = 10.5325[/tex]

You should claim 10.5325 years on your warranty.

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: Michelle's conclusion is wrong, he sells 8214.3 songs per day

Step-by-step explanation:

Sell 115000 songs every 2 weeks

Sell 115000 songs every 14 days

Sell (a) songs for 1 day

(a) =(115000x1) ➗ 14

(a) =115000 ➗ 14

(a) =8214.3

Answer:

9.9166667

Step-by-step explanation:

All of the numbers equal 59.5 then divide by 6 equals the answer above.

To calculate the mean of the given data set, sum all the data values and divide by the number of values. The mean of this data set is 9.9167.

To find the mean of the given data set: 2.5, 6.5, 9, 19, 20, 2.5,  follow these steps:

Hence, the mean of the data set is 9.9167.

Answer:

m1 = 62°, m2 = 118°

Step-by-step explanation:

The definition of a linear pair of angles is a pair of angles that are adjacent and add up to 180°.

Then, we have the following equation

[tex]x-9+x+47 =180 [/tex]

Then,

[tex]2x+38=180[/tex]

Subtracting 38 on both sides

[tex] 2x = 180-38 = 142[/tex]

Dividing by two, we get x = 71. Then m1 = 71-9 = 62, m2 = 71+47=118[/tex]