Apaint crew gets a rush order to paint 80 houses in a new development. they paint the first y houses at a rate of x houses per week. realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. the total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week? (a) 0.8(80 – y)(b) 0.8 + 0.0025y(c) 80/y – 1.25(d) 80/1.25y(e) 80 – 0.25y

(B) 0.8+0.0025y

Step-by-step explanation:

Total houses =80

First y houses was painted at the rate of x hours per week

Remaining houses was painted at 1.25x hours per week

Remaining houses= 80-y

Rate = quantity/ time

Time = quantity/ rate

Time for the first painting = y/x

Time for the second painting = 80-y/1.25x

Total time y/x + 80-y/1.25x

= 0.25y +80/1.25x

If it was being painted at the original rate

Time = 80/x

The time to paint in this scenario as a fraction of the time it will take to paint in the original rate.

(0.25y+80/1.25x) /(x/80)

=( 0.25y +80)/100

=0.0025y +0.8

answer: slight chance that it will happen

step-by-step explanation:

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]