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Roman55 [17]
3 years ago
15

1. Sharon will borrow $3400 at 11% APR. She will pay it back over 18 months. What will her monthly payment be? (Use table)

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0

Answer:

1) The Monthly payment is 220.06.

2) The Monthly payment is 73.24.

Step-by-step explanation:

1) Given :  Sharon will borrow $3400 at 11% APR. She will pay it back over 18 months.

To find : What will her monthly payment be?    

Solution : Amount is $3400 , APR = 11% , Time= 18 months  

From the table 1  we can see at 11% APR and in 18 months the interest per $100 is 16.50 (Marked in attached table 1)  

So, The total interest is  

I=16.50\times 34 = 561

The Monthly payment is  M= \frac{A+I}{t}        

M=\frac{3400+ 561}{18}=220.06        

Therefore,The Monthly payment is 220.06 Option B is correct.

2) Given : Patrick will borrow $4300 at 11.5% APR. He will pay it bak over 5 years.

To find : What will her monthly payment be?    

Solution : Amount is $4300 , APR = 11.5% , Time= 5 years

From the table 2  we can see at 11.5% APR and in 5 years the interest per $100 is 2.199 (Marked in attached table 2)  

So, The total interest is  

I=2.199\times 43 =94.557

The Monthly payment is  M= \frac{A+I}{t}        

M=\frac{4300+94.557}{5\times 12}=73.24          

Therefore,The Monthly payment is 73.24.

solong [7]3 years ago
5 0

Answer:

1.  The answer is 220.06

2.  Then the monthly payment for $4,300 would be: $4,300/100 x $2.199 =$94.56

Step-by-step explanation:


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Vedmedyk [2.9K]
y = \frac{1}{24}(x+1)^2 - 3\\\\y+3 =\frac{1}{24}(x+1)^2\ \ / *24\\\\ (x+1)^2 = 24(y+3)

This   is  an  equation  of  a  parabola  that  opens  upwards.

Its \ standard \ form: \\(x-h)^2=4p(y-k)\\ (h,k)=(x,y) \ coordinates \ of \ the \ vertex\\\ (h,k)=(-1,-3) \\\\axis \ of \ symmetry: \ x= -1\\ \\4p=24\ \ /:4\\p=6

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6 0
3 years ago
Read 2 more answers
10+4.5m=21.25<br> Solve for M
enyata [817]

<u>Answer:</u>

  • m = 2.5

<u>Step-by-step explanation:</u>

  • 10 + 4.5m = 21.25
  • => 4.5m = 21.25 - 10
  • => 4.5m = 11.25
  • => m = 11.25/4.5
  • => m = 2.5

<u>Conclusion:</u>

Therefore, m = 2.5

Hoped this helped.

GeniusUser

8 0
3 years ago
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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