Ask question

Ask question

All categories

3 years ago

11

Which represents the equation of a line passing through the points (-2, 3) and (5, 4)?

1 answer:

arsen [322]3 years ago

8 0

Answer:

y = (1/7)x + (23/7)

You might be interested in

5x-15=50 (Show your work)

Zigmanuir [339]

5x - 15 = 50
Add 15 to both sides
5x = 65
Divide both sides by 5
x = 13

Hope this helped :)

3 0

3 years ago

Read 2 more answers

Find the 66th term of the arithmetic sequence 25, 10, -5

siniylev [52]

Answer:

970

Step-by-step explanation:

8 0

4 years ago

3x2 + 5x - 7(x2 + 4)

goldenfox [79]

Answer:

-4x^2+5x-28

Step-by-step explanation:

Expand -7(x^2 +4) = 3x^2+5x-7x^2-28

Simplify 3x^2+5x-7x^2-28 = -4x^2+5x-28

5 0

3 years ago

Read 2 more answers

A doctor has an annual income of $152,125. The income tax the doctor has to pay is 6% , what is the amount of income tax in doll

kupik [55]

I believe the amount of income tax the doctor has to pay is $9,127.5

5 0

3 years ago

Read 2 more answers

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

a

$SE = 0.66}$

b

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago