What’s the answer please I need help.

John can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. How long does it take for both John and Rocky to

mario62 [17]

Answer: it will take them 48 minutes.

Step-by-step explanation:

John can mow a lawn in 80 minutes. This means that the rate at which he moans the lawn per minute is 1/80

Rocky can mow the same lawn in 120 minutes. This means that the rate at which Rocky can mow the same lawn per minute is 1/120

If they work together, they would work simultaneously and their individual rates are additive. This means that their combined working rate would be

1/80 + 1/120 = 200/9600 = 1/48

Assuming it takes t hours for both of them to clean the room working together, the working rate per hour would be 1/t. Therefore,

1/48 = 1/t

t = 48 minutes

4 0

3 years ago

Best Buy is having a sale. iPads are marked down 20%. Of the original cost is $399, how much will you pay?

iren [92.7K]

It is

$79.80 I believe
According to me?

5 0

2 years ago

Read 2 more answers

What is the solution to the system of equations?

lord [1]

Answer:

We are sorry, but you do not have access to this service. Please contact your Organization Administrator for access.

5 0

3 years ago

Three weeks ago John bought stock at 491/4; today the stock is valued at 497/8. We could say the stock is performing at which of

Feliz [49]

The correct answer on my test was Above Par

4 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