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neonofarm [45]
3 years ago
13

OK SO I NEED HELP WITH THIS:

Mathematics
1 answer:
White raven [17]3 years ago
6 0

9514 1404 393

Answer:

  g) $6000 ($528.18 to principal)

  h) $9,471.82

  i) $5471.82

Step-by-step explanation:

You can make a chart, as you have started. Such a chart is often referred to as an <em>amortization schedule</em>. The basic idea would be to show how each payment is distributed to interest and principal. The attached shows such a spreadsheet (with a few rows hidden). It is described in more detail below.

__

g) The total amount paid is 24 payments of $250 each, so $6000. These payments reduced his amount owed by $528.18. (We don't know if "toward his credit card" means "to the creditor" or "to pay his debt".)

__

h) The amount still owed is $9,471.82 (the ending balance after the 24th payment)

__

i) $5,471.82 went toward interest.

_____

<em>Spreadsheet details</em>

The first "Beg Bal" is the $10,000 owed. Each one after that is copied from the "New Bal" cell on the previous line.

Each "to interest" quantity is the product of the beginning balance and the monthly interest rate. The cell reference to the interest rate needs to be an absolute reference, so it remains the same when copying the formula to other cells.

The "to Principal" amount is the difference between the payment amount and the amount "to interest." The reference to the payment amount cell also must be an absolute reference.

The "End Bal" amount is the difference between the "Beg Bal" and the amount "to Principal".

As you can see, the payment math is not difficult. There is one multiplication (by the interest rate) and the rest is subtraction.

Note that the numbers in this spreadsheet are rounded by the display. In a real payment situation, the interest due amount would be rounded to the penny before being used in other calculations. The way we have done it will result in a few cents difference in the ending balance from the value if the interest were actually rounded at each step.

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Step-by-step explanation:

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Question 2 -of 15 Step 1 of 1No Time LimitA company manufactures two products. One requires 5 hours of labor, 3 poundsof raw mat
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the cost of labour per hour is $7.20

the cost of raw materials per pound is $11.60

Explanation:

For product one:

time = 5 hours of labour

let the cost labour per hour = x

Amount = 3 pounds of raw amterials

let the cost of one pound raw material = y

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The equation:

time (cost per hour) + amount (cost of one pound of raw material) = Cost to produce each product

5(x) + 3(y) = 70.80

5x+3y=70.8....\mleft(1\mright)

For product 2:

time = 3.5 hours of labour

let the cost of labour per hour = x

Amount = 13 pounds of raw amterials

let the cost of one pound raw material = y

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The equation:

time (cost per hour) + amount (cost of one pound of raw material) = Cost to produce each product

3.5(x) + 13(y) = 176

3.5x+13y=176\text{   .... (2)}

combining both equations:

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3.5x + 13y = 176 ....(2)

Using elimination method:

To eliminate y, we will multiply equation (1) by 13 and equation (2) by 3 so that both coefficient of y become the same

65x + 39y = 920.4 ...(*1)

10.5x + 39y = 528 ...(2*)

subtract equation (2*) from (1*):

65x - 10.5x + 39y - 39y = 920.4 - 528

54.5x + 0 = 392.4

54.5x = 392.4

divide both sides by 54.5:

x = 392.4/54.5

x = 7.2

substitute for x in any of the equations

Using equation 1: 5x + 3y = 70.8

\begin{gathered} 5\mleft(7.2\mright)+3y=\text{ 70.8} \\ 36\text{ + 3y = 70.8} \\ 3y\text{ = 70.8 - 36} \\ 3y\text{ = 34.8} \\  \\ \text{divide both sides by 3:} \\ \frac{3y}{3}=\frac{34.8}{3} \\ y\text{ = }11.6 \end{gathered}

Hence, the cost of labour per hour is $7.20 and the cost of raw materials per pound is $11.60

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Step-by-step explanation:

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