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Marta_Voda [28]
3 years ago
15

How long will it take to pay off a loan of ​$51000 at an annual rate of 12 percent compounded monthly if you make monthly paymen

ts of ​$650​? Use five decimal places for the monthly percentage rate in your calculations.
Mathematics
1 answer:
liraira [26]3 years ago
3 0

Answer:

Given,

The present value of the loan, P.V. = $ 51,000,

Annual rate of interest = 12 %,

So, the rate per month, r = \frac{12}{12} = 1 % = 0.01,

Also, the monthly payment, P = $ 650,

Let n be the number of months,

Since, monthly payment of a loan is,

P=\frac{P.V. (r)}{1-(1+r)^{-n}}

650=\frac{51000(0.01)}{1-(1+0.01)^{-n}}}

By graphing calculator,

n=154.299447759\approx 154.29944

Hence, it will take approximately 154 months to pay off the loan.

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Write the slope-intercept form of the equation that passes through the point (4,-6) and is parallel to the line y = -3/4x - 5
Karolina [17]

The equation that runs through the location (4,-6) has the slope-intercept form,  y=-\frac{3}{4}x-3 .

<h2>Formation of the equation</h2>

A line's equation written in the slope-intercept form:

y=mx+b

where m= slope & b= y-intercept

The slope of two parallel lines is equal.

Currently, we know the line's equation:

y=-\frac{3}{4} x-5

here, slope, m= -\frac{3}{4}

A line equation is created by adding the slope's value and the point's coordinates (4, -6):

-6=-\frac{3}{4}(4)+b

⇒ -6=-3 +b [adding 3 to both sides]

⇒-3=b

⇒b= -3

Hence the solution is  y=-\frac{3}{4}x-3 .

Learn more about slope-intercept form here:

brainly.com/question/9682526

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1 year ago
The model below represents the equation shown. What is the value of x?
Angelina_Jolie [31]

Answer:

Step-by-step explanation:

4 0
3 years ago
How is it going to tell me the answer
Anon25 [30]
I dont know man you gotta ask a question
7 0
3 years ago
Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow
prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

CI=\hat{Y}\pm t_{Critical}\times S_{e}

Where,

\hat{Y}= predicted value of lead content when traffic flow is 15.

\to df=n-1=8-1=7

 95\% \ CI\  is\  (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\

     =\frac{1059.8}{2}\\\\=529.9

Calculating thet-critical valuet_{ \{\frac{\alpha}{2},\ df \}}=-2.365

The lower predicted value =529.9-2.365(Se)

463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076

When 99\% of CI use as the expected lead content: \to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)

8 0
3 years ago
Guys please help its my final either i pass or not
Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

1 yes 2 no 3 no 4 yes 5 no

5 0
3 years ago
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