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sergiy2304 [10]
3 years ago
10

You have $2,000 on a credit card that charges a 16% interest rate. If you want to pay off the credit card in 5 years, how much w

ill you need to pay each month (assuming you don't charge anything new to the card)?
Mathematics
1 answer:
katrin2010 [14]3 years ago
8 0

9514 1404 393

Answer:

  $48.64

Step-by-step explanation:

The monthly payment amount is given by the amortization formula ...

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the loan amount, r is the annual interest rate compounded n times per year for t years.

Here, you have P=2000, r=0.16, n=12 (months per year), t=5 (years), so the payment is ...

  A = $2000(0.16/12)/(1 -(1 +0.16/12)^(-12·5)) = $320/(12(0.54828942))

  A ≈ $48.636 ≈ $48.64

You will need to pay $48.64 each month to pay off the charge in 5 years.

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Answer:

The probability is 0.4207

Step-by-step explanation:

The probability of a home-based computer having access to on-line services is p = 0.2 (data from the exercise)

Then, the probability of a home-based computer not having access to on-line services is p = 1 - 0.2 = 0.8

We are going to use this probability (p = 0.8) to solve the exercise.

Let's define the random variable X

X : ''Number of home-based computers not having access to on-line services''

X can be modeled as a binomial random variable

X ~ Bi(p,n)

X ~Bi(0.8,25)

Where p is the success probability and n is the number of Bernoulli independent experiments we are taking place.

We are going to count ''a success'' as a computer not having access to on-line services.

The binomial probability function is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as

nCx=\frac{n!}{x!(n-x)!}

p is the success probability and n the number of Bernoulli independent experiments taking place.

In our exercise,

p=0.8\\n=25

We are looking for :

P(X>20)=P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)

P(X>20)=(25C21)0.8^{21}0.2^{4}+(25C22)0.8^{22}0.2^{3}+(25C23)0.8^{23}0.2^{2}+(25C24)0.8^{24}0.2^{1}+(25C25)0.8^{25}0.2^{0}

P(X>20)=0.1867+0.1358+0.0708+0.0236+0.8^{25}

P(X>20)=0.4207

Finally, the probability of finding that more than 20 of 25 home-based computers do not have access to on-line services is 0.4207

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