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Rama09 [41]
3 years ago
10

What is a monthly finance charge if the average daily balance is $30, the daily periodic rate is 0.07% and the number of days in

a cycle is 30?
A. 21 cents
B. 63 cents
C. 90 cents
Mathematics
2 answers:
Wewaii [24]3 years ago
6 0

Answer:

The finance charge is:

Option: B

B. 63 cents.

Step-by-step explanation:

We are given:

Average daily balance = $30

The daily periodic rate = 0.07%

and Number of days in a cycle is 30.

We know that the formula for the finance charge is given as:

Finance\ charge=\dfrac{APR\cdot Balance\cdot Number\ of\ days\ in\ billing\ cycle}{365}

As we are given daily periodic rate this means that:

\dfrac{APR}{365}=0.07\%=0.0007

Hence, putting the value in the formula we have:

Finance\ charge=0.0007\times 30\times 30\\\\Finance\ charge=0.63

finance charge=$ 0.63

As we know that:

1 dollar=100 cents.

Hence, 0.63 dollars=63 cents

Hence, the finance charge is:

63 cents

Sloan [31]3 years ago
5 0

B. 63 Cents is the correct answer.

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
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Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

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Volume of tank = x³

The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = 64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

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