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Likurg_2 [28]
2 years ago
14

You decide that you want to purchase a Tesla SUV. You borrow \$95,000 for the purchase. You agree to repay the loan by paying eq

ual monthly payments of \$1,200 until the balance is paid off. If you're being charged 6\% per year, compounded monthly, how long will it take you to pay off the loan?
Mathematics
1 answer:
liraira [26]2 years ago
5 0

Using compound interest and a graphing calculator, it is found that it will take about 15 years for the loan to be paid off.

<h3>What is compound interest?</h3>

The amount of money earned, in compound interest, after t years, is given by:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

In which:

  • A(t) is the amount of money after t years.
  • P is the principal(the initial sum of money).
  • r is the interest rate(as a decimal value).
  • n is the number of times that interest is compounded per year.

For this problem, the parameters are given as follows:

A(0) = 95000, r = 0.06, n = 12.

Hence the value of the loan after t years is:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

A(t) = 95000\left(1 + \frac{0.06}{12}\right)^{12t}

A(t) = 95000(1.005)^{12t}

You have monthly payments of $1,200, hence the amount paid after t years is:

P(t) = 12 x 1,200t = 14400t

Then we have to solve for:

A(t) = P(t)

14400t = 95000(1.005)^{12t}

Which is solved in the graph below, meaning that it will take about 15 years for the loan to be paid off.

More can be learned about compound interest at brainly.com/question/25781328

#SPJ1

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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p
skelet666 [1.2K]

Answer:

\mathbf{P(X=5) =0.0888}    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

\mathtt{P(X=x) =(^{n}_{x} )   \  \pi^x \  (1-\pi)^{n-x}}

\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} )   \  \pi^x \  (1-\pi)^{n-x}}

where;

n = 8 and π = 0.36

For x = 5

The probability \mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} )   \  0.36^5 \  (1-0.36)^{8-5}}

\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} )   \  0.36^5 \  (0.64)^{3}}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =({8 \times 7 } )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =0.0887645}

\mathbf{P(X=5) =0.0888}     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})

{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times  (0.36)^0  \times  (1-0.36)^8  \ )  +  \dfrac{8!}{1!(7!)} \times  (0.36)^1  \times  (1-0.36)^7  \ +\dfrac{8!}{2!(6!)} \times  (0.36)^2  \times  (1-0.36)^6  \ +  \dfrac{8!}{3!(5!)} \times  (0.36)^3  \times  (1-0.36)^5 +  \dfrac{8!}{4!(4!)} \times  (0.36)^4  \times  (1-0.36)^4  \  +  \dfrac{8!}{5!(3!)} \times  (0.36)^5  \times  (1-0.36)^3  \ )

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

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Answer:

Charlie can buy 14 hats at the sale price

Step-by-step explanation:

Let

h ----> the sale price of the hat

x ---> the regular price of the hat

we know that

x=7h ----> equation A

so

Two hats at the regular price  is equal to multiply 2 by x

Multiply by 2 equation A both sides

2x=(7h)*2

2x=14h

2 hats at the regular price cost the same that 14 hats at the sale price

therefore

Charlie can buy 14 hats at the sale price

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