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3 years ago

PLEASE HELP ASAP ILL GIVE BRAINLIEST AND A THANKS!

Mathematics

2 answers:

svetoff [14.1K]3 years ago

7 0

Answer:

28.26 cm²

Step-by-step explanation:

3² = 9

9 x 3.14 = 28.26

S_A_V [24]3 years ago

5 0

Answer:

28.27cm would be your answer :)

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Sixty is what percent of 50? A. 30% B. 90% C. 110% D. 120%

Gnom [1K]

A 30% The answer is A because if you multiply 50 by .60 % your answer will be 30.

3 0

3 years ago

The table shows four numeric expressions. Which expression is an integer when it's evaluated?

Paraphin [41]

9 * -3 = -27, which is an integer.

The answer is (A).

6 0

3 years ago

Read 2 more answers

Helpppppppp pleaseee

valina [46]

Answer:

let me give you some advice. i think that you should try out GOOGLE perimeter calculator

6 0

3 years ago

A hospital recorded the weights, in ounces, of newborn babies for two weeks. The

Gwar [14]

Answer:

The standard deviation for week two was about 3 ounces more than the standard deviation for week one

Step-by-step explanation:

Given

$Week\ 1: 128, 105, 80, 82, 96, 98, 87, 100, 112, 126$

$Week\ 2: 75, 85, 90, 97, 89, 105, 110, 127, 129, 130$

See attachment for options

Required

The true statement

Checking the standard deviation

For week 1

Calculate the mean:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{128+ 105+ 80+ 82+ 96+ 98+ 87+ 100+ 112+ 126}{10}$

$\bar x = \frac{1014}{10}$

$\bar x_1 = 101.4$

Then standard deviation

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

$\sigma_1 = \sqrt{\frac{(128 - 101.4)^2 +............+ (126- 101.4)^2}{10}}$

$\sigma_1 = \sqrt{\frac{2522.4}{10}}$

$\sigma_1 = \sqrt{252.24$

$\sigma_1 = 15.88$

For week 2, we have:

$\bar x = \frac{75+ 85+ 90+ 97+ 89+ 105+ 110+ 127+ 129+ 130}{10}$

$\bar x = \frac{1037}{10}$

$\bar x_2 = 103.7$

Then standard deviation

$\sigma_2 = \sqrt{\frac{(75 - 103.7)^2 +................+ (130- 103.7)^2}{10}}$

$\sigma_2 = \sqrt{\frac{3538.1}{10}}$

$\sigma_2 = \sqrt{353.81$

$\sigma_2 = 18.81$

Compare the standard deviations

$\sigma_1 = 15.88$

$\sigma_2 = 18.81$

Calculate the difference:

$d = \sigma_2 - \sigma_1$

$d = 18.81 - 15.88$

$d = 2.93$

$d \approx 3$

This implies that option (b) is true

4 0

3 years ago

A poll asked for people's opinion on whether closing local newspapers would hurt civic life; 427 of 1005 respondents said it wou

kykrilka [37]

Answer:

A. 0.4249

B. 95% CI = 0.4248-0.031 < p < 0.4249+0.030

C. 80% CI = 0.4249-0.0198 < p < 0.4249+0.0198

D.95% CI is wider than 80% CI.

The more confidence you want, the wider the CI must be.

Step-by-step explanation:

Poll asked for people's opinion on whether closing local newspapers would hurt civic life; 427 of 1005 respondents said it would hurt civic life a lot.

A.The proportion of the respondents who said closing local papers would hurt civic life a lot. (round to three decimal places as needed)

427/1005 = 0.4249

B. The 95% confidence interval for the population proportion who believed closing newspaper would hurt civic life a lot. Assume the poll used a simple random sample (SRS).

Therefore, 1.96√ 0.4249×0.57/1005 = 0.0304

95% CI 0.4248-0.031 < p < 0.4249+0.030

C. 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot.

= 1.2820×√[0.4249×0.57/1005] = 0.0198

80% CI: 0.4249-0.0198 < p < 0.4249+0.0198

D. 95% CI is wider than 80% CI.

The more confidence you want, the wider the CI must be.

8 0

3 years ago