A survey by the Pew Research Center found that social networking is popular in many nations around the world. The file here contains the level of social media networking (measured as the percent of individuals polled who use social networking sites) and the GDP per capita based on purchasing power parity (PPP) for each of 25 selected countries. At the 0.001 level of significance, is there a significant linear relationship between GDP and social media usage?

Answer:

  ∠3 = ∠4 = 60°

Step-by-step explanation:

Angles 1 and 3 are "remote interior angles" with respect to angle 2, so ...

  ∠2 = ∠1 + ∠3

  120° = 60° + ∠3 . . . fill in known values

  60° = ∠3 . . . . . . . . . subtract 60°

__

Since two of the interior angles in the triangle ABC are 60°, the third one is also. The interior angle at B (supplementary to angle 2) is corresponding to ∠4, so has the same measure as angle 4.

  ∠4 = 60°

Answer:

3n + 2

Step-by-step explanation:

-n + 4n -4 + 6

3n + 2

Using the concept of stratified sampling, it is found that 10 people will be selected from those who responded yes.

In a stratified sample, the population is divided into groups, and the same number of elements of each group is surveyed.

In this problem:

A similar problem is given at https://brainly.com/question/24188753

To decide how many participants who answered 'yes' to include in a stratified random sample of 20, use the proportion of 'yes' answers out of 300 to calculate the sample from that stratum.

In a stratified random sample, the population is divided into groups, or strata, and a sample is taken from each group to ensure that each subgroup of the population is adequately represented. To determine the number of people that will be selected from those who responded yes in a stratified random sample, we need to find the proportion of 'yes' responses among the 300 respondents and then apply that proportion to the sample size of 20.

Assuming we know the exact number of people who responded 'yes', let's call that number 'Y'. The number of 'yes' responses in the stratified sample would then be (Y/300) * 20. Without the actual number of 'yes' responses, we cannot compute the exact number of people that should be selected from the 'yes' group. However, this formula demonstrates how you would calculate it once the value for 'Y' is known.

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Answer:

Used car have the highest range of 75

Step-by-step explanation:

Yeah the type of car that have the highest range In monthly sales.

we know that

The range is the difference between the highest and the lowest value

First we,

calculate the range in monthly sales for the new car

highest value=51

lowest value=20

range=51-20

range=31

Secondly,we

calculate the range in monthly sales for used car

highest value=87

lowest value=12

range=87-12

range=75

Answer:

My answer: "The car that had the largest range in monthly sales was a used. Through finding the range by subtracting the highest and lowest data points, i was able to find the range for used cars being 75, and for new cars was 31. 75 is larger then 31, so therefore the used cars have the largest monthly range in sales. "

Their sample answer: Sample Response: I subtracted the highest and lowest numbers. The range for new cars was 31. The range for old cars was 75. The range for used cars was much bigger."

Select all that you included in your explanation.

The range for new cars was 31.

The range for used cars was 75.

Subtract the highest and lowest numbers.

Step-by-step explanation:

edg2020

Answer:

1.53

Step-by-step explanation:

Find the attachment for explanation

The z-value corresponding to the probability of 0.937 using the standard normal distribution table is 1.53 and this can be determined by using the given data.

Given :

Assuming that the distribution is normal so the probability for being within the maximum limit is given by:

[tex]\rm P=1-\dfrac{6.3}{100}[/tex]

P = 0.937

Now, the z-value corresponding to the probability of 0.937 using the standard normal distribution table is 1.53.

Therefore, the correct option is D) 1.53.

For more information, refer to the link given below:

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Answer:

Here, we have:

P(5 days snow in this 8 days) = 8C5 x (0.85)^5 x (1 - 0.85)^3 = 0.084

P(6 days snow in this 8 days) = 8C6 x (0.85)^6 x (1 - 0.85)^2 = 0.238

P(7 days snow in this 8 days) = 8C7 x (0.85)^7  x (1 - 0.85)^1  = 0.385

P(8 days snow in this 8 days) = 8C8 x (0.85)^8 x (1 - 0.85)^0 = 0.272

Add up those above, then the probability that it will snow AT LEAST five of those days in February:

P = 0.084+ 0.238 + 0. 385 + 0.272 = 0.979

Hope this helps!

:)

Answer: -28

Step-by-step explanation: Since x is being divided by -4, to solve for x, multiply both sides of the equation by -4.

On the left side, the -4's will cancel

and on the right side, 7(-4) is -28.

So x = -28.

Please do not try to do this problem in your head.

Show the work that it takes to get x by itself.

Answer:

x= - 28

Step-by-step explanation:

-4/1[x/-4 = 7]

x = -28

Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,580​, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,117​, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

[tex]\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}[/tex]

For n=25 we have:

[tex]\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580[/tex]

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

[tex]\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117[/tex]

For samples of size 25 and 50, the mean of x bar is $63,800. The standard deviation of x bar is $1580 for a sample size of 25 and $1117 for a sample size of 50.

a. For samples of size 25, the mean of x bar is equal to the population mean, which is $63,800. The standard deviation of x bar is equal to the population standard deviation divided by the square root of the sample size. So, the standard deviation of x bar is $7900/sqrt(25) = $1580.

b. For samples of size 50, the mean of x bar is still $63,800. The standard deviation of x bar is $7900/sqrt(50) = $1117. Note that as the sample size increases, the standard deviation of x bar decreases.

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Answer: 320°

Step-by-step explanation:

This is a circle geometry.

The arc length of the circle is given to be 8πcm and the radius is 4.5cm.

Now the length of an arc of a circle is

Arc length = πr0°/180° or 2πr0°/360°

To find the angle 0° subtend at the center we equate the arc length with the formula and solve for 0°.Now we go

πr0°/180 = 8π, convert to a simple linear equal and solve for the angle.

πr0° = 8π × 180

0°. = 8π × 180

-----------

π × r

= 8 × 180. 8 × 180

-------- or ---------

9/2. 4.5

= 8 × 180 × 2

------------

9

= 8 × 20 × 2

= 320°

or 8 × 180/4.5

= 1440/4.5

= 320°

We have been given that a sprinkler is designed to rotate 360∘ clockwise, and then 360∘ counterclockwise to water a circular region with a radius of 11 feet. The sprinkler is located in the middle of the circular region. The sprinkler begins malfunctioning and is only able to rotate 225∘ in each direction.

We are asked to find the area of the sector to nearest square foot.

We will use area of sector formula to solve our given problem.

[tex]\text{Area of sector}=\frac{\theta}{360}\times \pi r^2[/tex], where,

[tex]\theta[/tex] = Central angle of sector,

[tex]r[/tex] = Radius.

For our given problem [tex]\theta = 225^{\circ}[/tex] and [tex]r=11[/tex].

[tex]\text{Area of sector}=\frac{225^{\circ}}{360^{\circ}}\times \pi (11)^2[/tex]

[tex]\text{Area of sector}=0.625\times 121\pi[/tex]

[tex]\text{Area of sector}=237.5829444277281137[/tex]

[tex]\text{Area of sector}\approx 238[/tex]

Therefore, the sprinkler can water approximately 238 square feet.

To find the area of the sector, we need to find the central angle, find the fraction of the circle covered by the sector, and then multiply it by the area of the entire circle with a radius of 11 feet.

To find the area of the sector, we need to find the central angle first. Since the sprinkler can only rotate 225∘ in each direction, the total angle covered is 225∘+225∘=450∘.

Next, we need to find the fraction of the circle covered by the sector. We can do this by finding the ratio of the central angle to the total angle of a circle, which is 360∘. This can be calculated as (450/360).

Finally, we multiply the fraction by the area of the entire circle with a radius of 11 feet, which is π(11)^2, to find the area of the sector.

Answer:

[tex] \bar X = \frac{62+63+68+72+79+80+83+93+94+95}{10}= 78.9[/tex]

[tex] |62-78.9| = 16.9[/tex]

[tex] |63-78.9| = 15.9[/tex]

[tex] |68-78.9| = 10.9[/tex]

[tex] |72-78.9| = 6.9[/tex]

[tex] |79-78.9| = 0.1[/tex]

[tex] |80-78.9| = 1.1[/tex]

[tex] |83-78.9| = 4.1[/tex]

[tex] |93-78.9| = 14.1[/tex]

[tex] |94-78.9| = 15.1[/tex]

[tex] |95-78.9| = 16.1[/tex]

[tex] MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n}[/tex]

And replacing we got:

[tex] MAD =\frac{16.9+15.9+10.9+6.9+0.1+1.1+4.1+14.1+15.1+16.1}{10}= 10.12[/tex]

And the best anwer is

10.12

Step-by-step explanation:

We have the following data given:

62 63 68 72 79 80 83 93 94 95

And we need to begin finding the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X = \frac{62+63+68+72+79+80+83+93+94+95}{10}= 78.9[/tex]

Now we can find the mean absolute deviation like this:

[tex] |62-78.9| = 16.9[/tex]

[tex] |63-78.9| = 15.9[/tex]

[tex] |68-78.9| = 10.9[/tex]

[tex] |72-78.9| = 6.9[/tex]

[tex] |79-78.9| = 0.1[/tex]

[tex] |80-78.9| = 1.1[/tex]

[tex] |83-78.9| = 4.1[/tex]

[tex] |93-78.9| = 14.1[/tex]

[tex] |94-78.9| = 15.1[/tex]

[tex] |95-78.9| = 16.1[/tex]

And finally we can find the mean abslute deviation with the following formula:

[tex] MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n}[/tex]

And replacing we got:

[tex] MAD =\frac{16.9+15.9+10.9+6.9+0.1+1.1+4.1+14.1+15.1+16.1}{10}= 10.12[/tex]

And the best anwer is

10.12

Answer:

82.8 centimeters

Step-by-step explanation:

Chandler has 828 millimeters of fabric.

Expressing as a ratio

[tex]\dfrac{1}{x}= \dfrac{10}{828}\\10x=828\\x=828\div 10\\x=82.8 cm[/tex]

Therefore, Chandler has 82.8 centimeters of fabric.

Answer:

82.8 centimeters of fabric

Step-by-step explanation:

1 centimeters= 10 millimeters;

828mm x 1cm ÷10 mm=82.8cm ;centimeters of fabric

The exercise is performed by conversion factor and a smaller unit is transferred to a larger unit that is centimeters.

Answer:

Step-by-step explanation:

if x= -2

and P(x)=x^4+4x^3+6x^2-8

then P(-2)=(-2)^4+4*(-2)^3+6*(-2)^2-8=16-32+24-8=0

so P(x)=(x+2)* Q(x)   and P(x) is divisible by x+2

Answer:

B

Step-by-step explanation:

You can see that the dashes on the number line are going up by 1/4 which is equal to 2/8 (when you multiply 1/4 by 2). At point C your already at 6/8 which is a little larger than 5/8 so when you do a little less than 6/8 you get to point B which is the best answer.

Answer:

260

Step-by-step explanation:

A = bh

Answer:

The minimum score required for the scholarship is 644.

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 = 488, \sigma = 111[/tex]

What is the minimum score required for the scholarship?

Top 8%, which means that the minimum score is the 100-8 = 92th percentile, which is X when Z has a pvalue of 0.92. So it is X when Z = 1.405.

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

[tex]1.405 = \frac{X - 488}{111}[/tex]

[tex]X - 488 = 1.405*111[/tex]

[tex]X = 644[/tex]

The minimum score required for the scholarship is 644.

Answer:

[tex]\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}[/tex]

[tex] 0.293 \leq \sigma^2 \leq 6.736[/tex]

And in order to obtain the confidence interval for the deviation we just take the square root and we got:

[tex] 0.541 \leq \sigma \leq 2.595[/tex]

Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim

Step-by-step explanation:

Data provided

1.9, 2.4, 3.0, 3.5, and 4.2

We can calculate the sample mean and deviation from this data with these formulas:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

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

And we got:

[tex]\bar X= 3[/tex]

s=0.903 represent the sample standard deviation

n=5 the sample size

Confidence=95% or 0.95

Confidence interval

We need to begin finding the confidence interval for the population variance is given by:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

The degrees of freedom given by:

[tex]df=n-1=5-1=4[/tex]

The Confidence level provided is 0.95 or 95%, the significance is then[tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical values for this case are:

[tex]\chi^2_{\alpha/2}=11.143[/tex]

[tex]\chi^2_{1- \alpha/2}=0.484[/tex]

And the confidence interval would be:

[tex]\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}[/tex]

[tex] 0.293 \leq \sigma^2 \leq 6.736[/tex]

And in order to obtain the confidence interval for the deviation we just take the square root and we got:

[tex] 0.541 \leq \sigma \leq 2.595[/tex]

Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim

Answer:

The 90% confidence interval for the true population proportion that feels that the current season is as good as, or better, than previous seasons is (0.2372, 0.3428).

Step-by-step explanation:

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

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

In which

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

For this problem, we have that:

[tex]n = 200, \pi = \frac{58}{200} = 0.29[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.29 - 1.645\sqrt{\frac{0.29*0.71}{200}} = 0.2372[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.29 + 1.645\sqrt{\frac{0.29*0.71}{200}} = 0.3428[/tex]

The 90% confidence interval for the true population proportion that feels that the current season is as good as, or better, than previous seasons is (0.2372, 0.3428).

Answer:

[tex]0.29 - 1.64\sqrt{\frac{0.29(1-0.29)}{200}}=0.237[/tex]

[tex]0.29 + 1.64\sqrt{\frac{0.29(1-0.29)}{200}}=0.343[/tex]

And the confidence interval for this case would be (0.237; 0.343).

Step-by-step explanation:

We can begin find the proportion estimated of responded that they felt that quality standards have been maintained with the following formula:

[tex]\hat p = \frac{X}{n}[/tex]

And replacing we got:

[tex] \hat p =\frac{58}{200}= 0.29[/tex]

The confidence interval is given by 90%, and the significance level would be [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]

The confidence interval for the true proportion is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

Replacing the values we got:

[tex]0.29 - 1.64\sqrt{\frac{0.29(1-0.29)}{200}}=0.237[/tex]

[tex]0.29 + 1.64\sqrt{\frac{0.29(1-0.29)}{200}}=0.343[/tex]

And the confidence interval for this case would be (0.237; 0.343).

Answer:

She drove [tex]\frac{147}{2} miles/ hour[/tex]

Step-by-step explanation:

We are given that Lydia drove 441 miles in 6 hours.

We are supposed to find how fast did she drive in miles per hour

Distance covered by Lydia in 6 hours = 441 miles

Distance covered by Lydia in 1 hour =[tex]\frac{441}{6}[/tex]

Distance covered by Lydia in 1 hour =[tex]\frac{147}{2} miles/ hour[/tex]

Hence She drove [tex]\frac{147}{2} miles/ hour[/tex]

Given Information:

Probability of super event = P(S) = 0.0023

Number of suppliers = n = 3

Probability of unique event = P(U) = 0.014

Required Information:

Probability that all three suppliers will be disrupted = ?

Answer:

P(3) = 0.0023

Step-by-step explanation:

We want to find out the approximate probability that all three suppliers will be disrupted at the same time at some point during the next five years.

The required probability is given by

P(n) = P(S) + (1 - P(S))*P(U)ⁿ

Where P(S) is the probability of super event that will disrupt all suppliers, P(U) is the probability of unique event that would disrupt one of the suppliers and n is the number of suppliers.

P(3) = 0.0023 + (1 - 0.0023)*(0.014)³

P(3) = 0.0023 + (0.9977)*(0.014)³

P(3) = 0.0023

The correct option is C = 0.0023

Therefore, there is 0.23% probability that all three suppliers will be disrupted at the same time at some point during the next five years.

Answer:

8.1 moles

Step-by-step explanation:

Given parameters: Mass of water to be decomposed = 29.2g Unknown: Number of moles of oxygen. Solution: To solve this problem, we first write the balanced reaction equation :                 2H₂O    →     2H₂    +    O₂ Now convert the given mass of the water to number of moles;   Number of moles of water =      Molar mass of water = 2(1) + 16 = 18g/mol   Number of moles of water =   = 16.2moles     From the balanced reaction equation:     2 moles of water produced 1 mole of oxygen gas;     16.2 mole of water will produce     = 8.1moles of oxygen gas

Hope this helps you :3

Answer:

1/4 of his income.

Step-by-step explanation:

If Wilbur spends 2/3 of his income, 1/3 or 4/12 of it is left for other purposes (It is easier if everything has a common denominator of 12). And if he shares 1/12 of that remaining amount, there is 3/12 left. And when we simplify 3/12, we get 1/4.

*Mark me brainliest!*

25

Step-by-step explanation:

25

Answer:

  y = x^2 +8x +16

Step-by-step explanation:

t can be written in terms of x, then substituted into the equation for y.

  x = t -4

  x + 4 = t

  y = t^2 = (x +4)^2

  y = x^2 +8x +16

Answer:

x= 3,1

Step-by-step explanation:

-b ±  √b²-4(ac)/2a

4 ±  √(-4)² - 4 · (1·3)/2·1

x = 2 ± 1

x = 3,1

Answer:

1/4

Step-by-step explanation:

2/4 ÷ 2

Copy dot flip

2/4 * 1/2

We can cancel the 2 in the numerator and denominator

1/4 * 1/1

1/4

Answer:

1/12

Step-by-step explanation:

hope this helps you

Final answer:

The probability of rolling a 2 on a single 12-sided die is 1/12, which is approximately 8.33%.

Explanation:

If a single 12-sided die is tossed once, the probability of rolling a 2 is calculated by dividing the number of ways to roll a 2 by the total number of possible outcomes on the die. Since there is only one way to roll a 2, and there are 12 different possible outcomes on a 12-sided die, the probability is calculated as follows:

Count the number of favorable outcomes for rolling a 2: There is 1 way to roll a 2.

Count the total number of possible outcomes on a 12-sided die: There are 12 possible outcomes (1, 2, 3, ... 12).

Divide the number of favorable outcomes by the total number of possible outcomes to get the probability: P(rolling a 2) = 1/12.

Therefore, the probability of rolling a 2 on a 12-sided die is 1/12, which can also be expressed as approximately 8.33%.

Plugging in the values and calculating, we find that the 98% confidence interval is approximately (0.5847, 0.6553).

To construct a 98% confidence interval for the proportion of all American females who support the death penalty for convicted murders, we can use the formula:

CI = p ± z * √(p * (1-p) / n)

Where:

Using a standard normal distribution table or a calculator, we can find that the z-score for a 98% confidence level is approximately 2.33.

Plugging in the values into the formula:

CI = 0.62 ± 2.33 * √(0.62 * (1-0.62) / 3100)

Calculating the values:

CI = 0.62 ± 2.33 * √(0.62 * 0.38 / 3100)

CI = 0.62 ± 2.33 * √(0.235 / 3100)

CI = 0.62 ± 2.33 * 0.01516

CI = 0.62 ± 0.03535

CI ≈ (0.5847, 0.6553)

Therefore, the 98% confidence interval for the proportion of all American females who support the death penalty for convicted murders is approximately (0.5847, 0.6553).

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Answer:

The 99% of confidence limits for the proportion that plan to vote for the incumbent.

(0.6473 ,0.7527)

Step-by-step explanation:

Explanation:-

Given data the election of a local construction union involves 2,000 union members. Among them, 500 members are randomly selected.

Given large sample size 'N' = 2000

Given sample size 'n' = 500

Given data Of the 500 surveyed, 350 said they would vote for the incumbent.

The sample Proportion

                        [tex]p = \frac{x}{n} = \frac{350}{500} =0.7[/tex]

                       q = 1-p = 1 - 0.7 = 0.3

Confidence intervals:-

The 99% of confidence intervals are determined by

[tex](p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p+Z_{\alpha }\sqrt{\frac{pq}{n} } )[/tex]

The z- score of 0.99 level of significance =2.576

[tex](0.7-2.576\sqrt{\frac{0.7X0.3}{500} } , 0.7+2.576\sqrt{\frac{0.7X0.3}{500} } )[/tex]

on using calculator, we get

(0.7 - 0.0527 ,0.7+0.0527)

(0.6473 ,0.7527)

Conclusion:-

The 99% of confidence limits for the proportion that plan to vote for the incumbent.

(0.6473 ,0.7527)