All categories

2 years ago

Rashad earned $662 by working five days a week at his full time job what average pay per day did he earn from full time work

Mathematics

1 answer:

ziro4ka [17]2 years ago

5 0

Answer: $132.40

Step-by-step explanation:

To find the average, take the total amount of money, then divide it by the number of days he worked. So he worked 5 days a week, and got a total of $662, so divide. 662/5 to get $132.40.

Hope that helps :)

best of luck <3

You might be interested in

Can someone tell me if what I did in this statistics problem is correct or not and specially why please?

worty [1.4K]

Answer: ya correct

Step-by-step explanation:

4 0

2 years ago

What’s an irrational number between 5.2 and 5.5

Reika [66]

5.3851648071..... is the irrational number your looking for

3 0

2 years ago

the entry ticket for a baseball game costs $5 for an adult ticket and $3 for a child.A group of 20 people paid total entry fees

sesenic [268]

There would be 15 adults and 5 children in that group.
15 adults = $75
5 Children= $15
75+15= $90

7 0

3 years ago

Read 2 more answers

And English class has read the first 384 pages of your novel. If this is 75% of the entire book, how many total pages are in the

Doss [256]

384 pages (0.75) = 288 pages
Add this to the original amount of pages.

384 + 288 = 672 total pages.

3 0

2 years ago

Read 2 more answers

A research company desires to know the mean consumption of meat per week among people over age 40. A sample of 610 people over a

Papessa [141]

Answer: $(3.0,\ 3.2)$

Step-by-step explanation:

Confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$ (1)

, where $\overline{x}$ = Sample mean

$z_{\alpha/2}$ = Critical z-value

$\sigma$ = Population standard deviation.

n= Sample size.

As per given , we have

n= 610

$\sigma=1.1$

$\overline{x}=3.1$

Significance level for 85% confidence : $\alpha=1-0.85=0.15$

By z-table critical two tailed z-value : $z_{\alpha/2}=z_{0.075}=1.44$

Put all values in (1) , we get

$3.1\pm 1.44\dfrac{1.1}{\sqrt{610}}$

$3.1\pm 1.44\dfrac{1.1}{24.698}$

$3.1\pm 0.064$

$=(3.1-0.064,\ 3.1+0.064)=(3.036,\ 3.164)\approx(3.0,\ 3.2)$

Hence, the 85% confidence interval for the mean consumption of meat among people over age 40. = $(3.0,\ 3.2)$

5 0

3 years ago