In which of the following situations does Alex's hourly wage change by a constant percent per unit change in year?

A. Alex's starting hourly wage is $14.50 per hour the first year, and it increases by $1.50 each year.

B. Alex's starting hourly wage is $13.00. She receives a $0.50 per hour raise after one year, a $0.75 per hour raise after the second year, a $1.00 per hour raise after the third year, and so on.

C. Alex's hourly wage is $20 per hour in the first year, $22 per hour the second year, $24.20 per hour the third year, and so on.

D. Alex's starting hourly wage is $15.00. Her hourly wage is $15.75 after one year, $17.00 after two years, $18.75 after three years, and so on.

 

I think the answer is A, can you correct me if I am wrong.

in Algebra 1 Answers by Level 1 User (220 points)

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1 Answer

Let's look at each answer in turn.

A. It can't be the answer. The reason is that the constant amount increases the wage year on year, so the wage is getting bigger but the increment is becoming a smaller fraction, or percentage, of the wage as time goes on.

B. This can't be the answer even though the increment itself is increasing. The percentage increase isn't in step with this. 0.50/13.50=3.7% approx. The wage is $14/hr. Next year the percentage is 0.75/14=5.4%, so the percent increase is not constant.

C. This is the correct answer because the wage rises by $2, which is 2/20=10% increase. Next year the increase percent is 2.20/22=10%. That's a constant percent increase.

D. This can't be correct if C is correct. Let's see what we've got. 0.75/15=5%; 1.75/17=10.3%. Not a constant increase.

by Top Rated User (1.2m points)
Solve B= -2.58w + 20 for w

Add 2.58w to each side: B+2.58w=20; subtract B from each side: 2.58w=20-B.

So w=(20-B)/2.58. This cannot be "solved" as a numerical value until B is known as a number, so this answer is simply making w the subject of the equation.

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