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2 years ago

Could someone check this. I don't know if I'm correct. Thanks!

Mathematics

1 answer:

garik1379 [7]2 years ago

8 0

Answer:

yes

Step-by-step explanation:

your answers are correct! congrats!

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Which ordered pairs make both inequalities true?

Arlecino [84]

Answer:

(0,-2) and (1, 1)

Step-by-step explanation:

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3 years ago

Solve this please<br><img src="https://tex.z-dn.net/?f=%20%20logx%20%2B%20%20log%20%5Cfrac%7Bx%20-%202%7D%7B21%7D%20%3D%203%20lo

VLD [36.1K]

Answer:

x = 14 or —12

Step-by-step explanation:please see attachment for explanation

5 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

Calculate the capacity in liters of a tin 20m by 15m by 10m.

rusak2 [61]

20 * 15 * 10 *10^-6
= 0.003

5 0

3 years ago

Please help I'm on a test right now!!

Svetach [21]

The answer would be 7-x.

8 0

2 years ago

Read 2 more answers

Could someone check this. I don't know if I'm correct. Thanks!​

Could someone check this. I don't know if I'm correct. Thanks!