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lidiya [134]
2 years ago
6

A certain business produced x rakes each month form November through February and shipped x/2 rakes at the beginning of each mon

th from March through October. The business paid no storage costs for the rakes from November through February, but it paid storage costs of $0.10 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form November through October?
A. 0.40x
B. 1.20x
C. 1.40x
D. 1.60x
E. 3.20x
Business
1 answer:
Natasha_Volkova [10]2 years ago
5 0

Answer:

C. 1.40x

Explanation:

1. Production:

November through February: x rakes/month* 4 months = 4x rakes.

2. Shipping and storage costs

March= 4x (initial stock) - x/2 (shipped) = 3.5 x (in stock) * 0.1$ = $0.35

April = 3.5x (stock at end of March) - x/2 (shipped) = 3x*0.1$ =  $0.30

May = 3x (stock at the end of April) - x/2 (shipped) = 2.5x*0.1$ = $0.25

June= 2.5x (stock at the end of May) - x/2 (shipped) = 2x (in stock) * $0.1 = 0.2$

July = 2x (stock at end of June) - x/2 (shipped) = 1.5x*0.1$ =  $0.15

August = 1.5x (stock at the end of July) - x/2 (shipped) = 1x*0.1$ = $0.1

Sept= 1x (stock at the end of August.) - x/2 (shipped) = 0.5 x  (in stock) * $0.1 = $0.05

October = 0.5x (stock at end of sept.) - x/2 (shipped) = 0*$0.1 =  $0

Total storage cost= $0.35x+$0.30x+$0.25x+$0.20x+$0.15x+$0.10x+$0.05x+$0x

Total storage cost = $1.40X

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6 0
3 years ago
Cindy invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. approximately how much does she
stiks02 [169]

Annual Compound Formula is:

A = P( 1 + r/n) ^nt

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r is the annual interest rate

<span>n is the number of  interest compounded per year</span>

t is the number of years the money is invested


So for the given problem:

P = $10,000

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snow_lady [41]

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The computation of the future value is shown below:

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Future value = Present value × (1 + interest rate)^number of years

In the first case,

Future value = $2,050 × (1 + 0.12)^12

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In the third case,

Future value =  $72,355× (1 + 0.11)^13

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In the fourth case,

Future value = $179,796 × (1 + 0.07)^7

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Answer:

True

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5 0
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