Ask question

Ask question

All categories

3 years ago

15

PLEASE HELP!!!!!

1 answer:

nasty-shy [4]3 years ago

6 0

The total charge would be $445.

Hope this helps!!

You might be interested in

Pls answer these questions correctly for brainly

FinnZ [79.3K]

40 - 3k + 10k = -10 - 8k - 10

40 + 7k = -10 - 8k - 10

40 + 7k = -8k - 20

40 + 7k - 40 = -8k - 20 - 40

7k = -8k - 60

7k + 8k = -8k - 60 + 8k

15k = -60

15k/15 = -60/15

k = -4

8 0

3 years ago

9. If 15 men working 10 days earn $500. How much will 12 men earn working 14,

Leno4ka [110]

Answer:

Firstly need to understand the questions.

If 15 men working 10 days, They earn 500.

If 12 men working 14 days, how much they will earn?

Step-by-step explanation:

560

Solution:

15 men working for 10 days =500

1 man working for 1 day = 500/(10*15) =10/3

Now 12 men working for 14 days = 12*14*10/3 =560.

I am pretty sure that answer is right.

Thank you :)

6 0

3 years ago

A 25 kg rock , is raised to the top of a hill 10 m high. What is the PE of the rock ?

nexus9112 [7]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$Potential \: \: Energy =(mass) \times (gravity \: acceleration) \times (height) \\$

_________________________________

If gravity acceleration is 10 :

$PE = 25 \times 10 \times 10$

$PE = 2500$

_________________________________

If gravity acceleration is 9.8 :

$PE = 25 \times 9.8 \times 10$

$PE = 25 \times 98$

$PE = 2450$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

3 0

3 years ago

natka813 [3]

Answer:

21s+8=r

need more characters

6 0

3 years ago

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