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3 years ago

∞•PLEASE HELP NOW IT'S URGENT AND YOU GET 10 POINTS•∞

Mathematics

1 answer:

Kamila [148]3 years ago

7 0

Answer:

they Distance between the two is 2.8

Step-by-step explanation:

d= (−1 −1 ) 2 + ( −1−1 ) 2

d= (−2 ) 2+ ( −2 ) 2

d= 4 + 4

d= 8–

d= 2.8

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Nine to the yth power?

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Answer:

9^y

Step-by-step explanation:

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3 years ago

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Simplify the square root of 7+2<br> So 7+2 is inside the quare root

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Answer:

Step-by-step explanation:

$\sqrt{7+2}=\sqrt{9}=\sqrt{3^2}=\boxed{3}$

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3 years ago

How would you change 2x^2=8x-3 into an equation you could solve using the quadratic formula

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3 years ago

A Gallup Poll used telephone interviews to survey a sample of 1030 U.S. residents over the age of 18 regarding their use of cred

Vlada [557]

Answer:

The 95% margin of error for this estimate is 0.0274 = 2.74 percentage points.

Step-by-step explanation:

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

72% of Americans said that they had at least one credit card

This means that $p = 0.72$

Give the 95% margin of error for this estimate.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{0.72*0.28)}{1030}} = 0.0274$

The 95% margin of error for this estimate is 0.0274 = 2.74 percentage points.

7 0

4 years ago

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,

Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ ( $z_{\alpha /2$ × σ ) / E ]²

given that confidence level = 95%, so $z_{\alpha /2$ = 1.96

Now,

a) Margin of error E = 500

n = [ ( $z_{\alpha /2$ × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ ( $z_{\alpha /2$ × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c) Margin of error E = 100

n = [ ( $z_{\alpha /2$ × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0

3 years ago