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EleoNora [17]
3 years ago
13

According to the Bureau of Labor Statistics, students who have an after-school job spend an average of 42 fewer minutes per day

on schoolwork than students who don't work. If the school year is 180 days, how many more hours per year would the average non-working student spend on schoolwork than the average student with an after-school job?
Mathematics
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer:

126 hrs

Step-by-step explanation:

-Given the school year has 180 days and the non-working student spends 42 minutes more than the working student.

-The total time during the school year is calculated as;

Total Time=Time/day \times No \ of \ Days\\\\=42\  minutes \times 180\\\\=7560\ minutes

We convert our calculated time in hour(1hr=60 minutes):

=\frac{7560\ mins}{60\ mins}\\\\=126\ hrs

Hence, the average non-working spends an average of 126 hrs that the working student.

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A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo
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Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the boxV=x^2z

Volume of the box=12 ft^3\\

Therefore, x^2z=12\\z=\frac{12}{x^2}

The material for the base costs \$0.17/ft^2, the material for the sides costs \$0.10/ft^2, and the material for the top costs \$0.13/ft^2.

Area of the base =x^2

Cost of the Base =\$0.17x^2

Area of the sides =4xz

Cost of the sides==\$0.10(4xz)

Area of the Top =x^2

Cost of the Base =\$0.13x^2

Total Cost, C(x,z) =0.17x^2+0.13x^2+0.10(4xz)

Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

To minimize C(x), we solve for the derivative and obtain its critical point

C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2

Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

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