Solve express the answer in simplest form. 1/5+w=7 1/3 w=__ _/

If the revenue for the week is $2000 and the labor cost consists of two workers earning eight dollars per hour who work 40 hours

arsen [322]

To find the cost of labor for one worker, you multiply the wage by the number of hours worked.

40 * 8 = 320

Because there are two workers, now you double it.

320 * 2 = 640

To find the percentage of the revenue that this is, you do the labor cost divided by the revenue.

640 / 2000 = 0.32

Move the decimal two places to the right to get the percentage.

0.32 = 32%

The labor cost is 32% of the revenue.

8 0

4 years ago

Can someone help me please asap

riadik2000 [5.3K]

Answer:

A. 55°

B. 71°

C. 100°

D. 65°

E. 120°

F. 155°

G. 147°

H. 36°

I. 70°

J. 54°

K. 24°

L. 40°

Step-by-step explanation:

A. 180 - (90 + 35) = 55°

B. 180 - (34 + 75) = 71°

C. 180 - (43 + 38) = 100°

D. 180 - 50 = 130 / 2 = 65°

E. 180 - (90 + 30) = 60°

180 - 60 = 120°

F. 180 - (120 + 35) = 25°

180 - 25 = 155°

G. 180 - 145 = 35 + 112 = 148°

H. 180 - (40 + 26) = 144°

180 - 144 = 36°

I. 180 - (60 + 65) = 125°

180 - 125 = 55°

180 - (55 + 50) = 75°

180 - (75 + 35) = 70°

J. 180 - 116 = 64°

180 - (64 + 62) = 54°

K. 180 - (138 +18) = 24°

L. 180 - (80 + 60) = 40°

7 0

3 years ago

Solve for x : ax+v=3(x-c)

never [62]

Answer:

$x=\frac{3c+v}{a-3}$

Step-by-step explanation:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

3 0

4 years ago

Read 2 more answers

Find the solution to the equation 16.2 + n = 16

kozerog [31]

The answer is N=-0.2

4 0

3 years ago

Read 2 more answers

. You are interested in the amount of time teenagers spend weekly working at part-time jobs. A random sample of 15 teenagers was

agasfer [191]

Answer:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

$\bar X=147.3$ represent the sample mean for the amount of time spent at part time jobs

$s=50$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =120$ represent the value to check

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

4 0

3 years ago

Solve express the answer in simplest form. 1/5+w=7 1/3 w=__ _/_