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3 years ago

7

What is ( 12 ) divided by 3??

1 answer:

GarryVolchara [31]3 years ago

4 0

12 divides by 3 is equal to 4

You might be interested in

ABCD is a rectangle. Find the length of each diagonal.ac=2(5a+1) bd=2(a+1).

Igoryamba

Rectangle diagonals are equal.

2(5a+1) = 2(a+1)
10a + 2 = 2a + 2
10a -2a = 2 - 2
8a = 0
⇒ a = 0

AC = 2(5a+1) = 2(5 × 0 +1)= 2(0 + 1) = 2 × 1 = 2 <span>
</span>BD = 2(a+1) = 2(0 + 1) = 2 <span>× 1 = 2
</span>
Ansver: AC = BD = 2

7 0

3 years ago

Jasmine by 32 kiwi fruit for $16.How many kiwi can lisa buy if she has $6

zimovet [89]

The way to solve this is take 32/16, and you get 2. This means that you get 2 kiwis per dollar. So, if Lisa has $6, then she can buy 12 kiwis.

5 0

2 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden

solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99% = 1 - 0.99

α = 0.01

and $\frac{\alpha}{2}$ = $\frac{0.01}{2}$

$\frac{\alpha}{2}$ = 0.005

so Z is here

$Z_(\frac{\alpha}{2})$ = 2.576

so

sample size will be

Sample size n = $(\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)$

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0

3 years ago

Find m∠PQR. A. 81 B. 90 C. 77 D. 72

AnnZ [28]

<h3>Answer: D. 72</h3>

Work Shown:

Angle PQR = (far arc - near arc)/2

Angle PQR = ( (major arc PR) - (minor arc PR) )/2

Angle PQR = ( (2x+252) - (2x+108) )/2

Angle PQR = 144/2

Angle PQR = 72

Notice how we didn't need to find the value of x at all.

5 0

2 years ago