Ask question

Ask question

All categories

3 years ago

10

I need help!! What is he answer??

1 answer:

IgorC [24]3 years ago

7 0

2 is the answer i think

You might be interested in

How do you solve 8x-3y=-70 and -2x+5y=-8 using the elimination method please show work.

lora16 [44]

Answer:

6x+2y=-78

Step-by-step explanation:

8x-3y=-70

-2x+5y=-8

6x+2y=-78

3 0

3 years ago

Read 2 more answers

Can someone help me with number 4?

Rashid [163]

The US: 300m World: 9b Most accurate: B

300m•20=6b population

B is the most accurate

4 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%20g%28x%29%3D%5Cfrac%7B1%7D%7Bx%7D%20%20" id="TexFormula1" title=" g(x)=\frac{1}{x} " alt

Ksju [112]

Fill in the given values and simplify.

$\dfrac{g(x+h)-g(x)}{h}=\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\\\\=\dfrac{x-(x+h)}{h\cdot x(x+h)}\\\\=\dfrac{-1}{x(x+h)}$

4 0

3 years ago

10>29-3b Please help and explain your answer I will mark Brainlyest

Marta_Voda [28]

Step 1 , switch sides - 29-3b<10
step 2 , subtract 29 from both sides - -3b< -19
step 3 , multiply both sides by -1 - 3b > 19
step 4 (answer) - divide by 3 = 19/3

7 0

3 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