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3 years ago

HELP ASAP An office employee ears $600 weekly. Find the equivalent earnings if paid biweekly, semimonthly, monthly, and annually

Mathematics

1 answer:

beks73 [17]3 years ago

6 0

1200~bi-weekly

1200~semimonthly

2400~monthly

31,200~annually

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RES Find (fºg)(1) Ax) = x2 - 1 g(x)= v a 0 c. 4. b. d. 18 Help pls

slava [35]

Answer:

Step-by-step explanation:

So we have the two functions:

$f(x)=x^2-1\\g(x)=\sqrt x$

And we want to find:

$(f\circ g)(1)$

This is the same as:

$f(g(1))$

So, to find the answer, find g(1) first:

$g(x)=\sqrt x\\g(1)=\sqrt1\\g(1)=1$

Now, substitute this value for g(1):

$f(g(1))\\=f(1)$

And plug this into f(x):

$f(x)=x^2-1\\f(1)=(1)^2-1\\f(1)=1-1=0$

Therefore:

$f(1)=f(g(1))=(f\circ g)(1)=0$

Our answer is A

3 0

4 years ago

Read 2 more answers

You bought a car for $19,999 and your bank requires 10% down. How much money is

valkas [14]

Answer:

$17,999.10

Step-by-step explanation:

Let x represent the amount the bank will lend.

x = $19,999(1 -10%) = 0.90($19,999) = $17,999.10

The bank is willing to lend $17,999.10 on the car.

6 0

4 years ago

How will increasing the level of confidence without changing the sample size affect the width of a confidence interval for a pop

atroni [7]

Answer:

The width of the confidence interval is going to be wider

Step-by-step explanation:

When we Increase the level of confidence without changing the sample size the width of the confidence interval is going to be affected in the following way;

The margin of error (M.E) is going to be larger or be increased. This increase is due to the fact that the critical value is going to increase. Then the increased margin of error would cause the width of the confidence interval to become wider.

This answers your question.

3 0

3 years ago

In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one

Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

$\text{Total ways}=^9C_3\times ^1C_1$