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Anna007 [38]
2 years ago
15

For 1≤n≤10 1 ≤ n ≤ 10 , find a formula for pn, the payment in year n on a loan of 105000 dollars. Interest is 3 percent per year

, compounded annually, and payments are made at the end of each year for 10 years. Each payment is 10500 dollars plus the interest on the amount of money outstanding.
Mathematics
1 answer:
Softa [21]2 years ago
3 0
I don’t know the answer but if I don’t answer this question I couldn’t answer my question
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Answer: the answer is letter A

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I say that x equals 5
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Each box is shaped like a rectangular prism. Which has more more storage space? Explain. Show all your work.
V125BC [204]

Answer:

Box A

Step-by-step explanation:

Find the volume of each one:

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Box B: 3*4*5= 60

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8 0
3 years ago
. Consumer Awareness In 1986, the average cost of a new midsize four-door sedan was $9000. In 1991, the average cost was $12,000
eduard

<u>Answer-</u>

\boxed{\boxed{y=9000(1.0592)^x}}

<em>And cost of a car in the year 2000 will be </em><em>$20134.18</em>

<u>Solution-</u>

Let's assume,

x = number of years after 1986

y = average cost of sedan in dollar

The exponential model that will model the scenario will be in the form of,

y=ab^x

where a and b are constants

As given that, in 1986 average cost of sedan was $9000 and in 1991 average cost of sedan was $12000

So, the points (0, 9000) and (5, 12000) will satisfy or lie on the exponential curve.

Putting (0, 9000) in the equation,

\Rightarrow 9000=ab^0

\Rightarrow a\times 1=9000

\Rightarrow a=9000

Now, the equation becomes y=9000b^x

Putting (5, 12000) in this equation,

\Rightarrow 12000=9000b^5

\Rightarrow b^5=\dfrac{12000}{9000}

\Rightarrow b=\sqrt[5]{\dfrac{4}{3}}

\Rightarrow b=1.0592

Putting the values,

y=9000(1.0592)^x

As we have to calculate the cost of sedan in 2000, so putting x=14(as 2000-1986=14),

y=9000(1.0592)^{14}=20134.18

Therefore, cost of a car in the year 2000 will be $20134.18

8 0
2 years ago
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