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2 years ago

Which is an equation of the line through the origin and (3,4)? Please help

Mathematics

1 answer:

kondor19780726 [428]2 years ago

8 0

B is the answers………..

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What is/are the solution(s) of the system of equations? Why you say that is the solution? Write your answer in complete sentence

Alexus [3.1K]

Answer: The answer to a system of equations is the point(s) of intersection of the equation involved.

When you have more than one equation with more than one variable, you have a system of equation. The solution(s) is the set of values for each variable that satisfy the equations. This means that the inputting values make the equations true.

6 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

A = 40, c = 41, b =<br><br> help please

nikitadnepr [17]

Answer:

I'm thinking it's 42. Did you notice that there is a pattern?

4 0

3 years ago

382 3/10 - 191 87/100=

frosja888 [35]

$382 \frac{3}{10} - 191 \frac{87}{100} =(382 + \frac{3}{10} ) - (191 + \frac{87}{100}) =\\\\=382 + \frac{30}{100} -191- \frac{87}{100} =382 -191+ \frac{30}{100}- \frac{87}{100} =\\\\=191+ \frac{30}{100}- \frac{87}{100} =190+ \frac{100}{100}+ \frac{30}{100}- \frac{87}{100} =\\\\=190+\frac{130}{100}- \frac{87}{100} =190+\frac{43}{100}=190\frac{43}{100}\\\\\\382 \frac{3}{10} - 191 \frac{87}{100} =382.3 - 191.87 =382.3 - 191-0.87 =191.3-0.87=\\\\=190+1.3-0.87=190+0.43=190.43$