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xeze [42]
3 years ago
13

Niki makes the same payment every two months to pay off his $61,600 loan. The loan has an interest rate of 9.84%, compounded eve

ry two months. If Niki pays off his loan after exactly eleven years, how much interest will he have paid in total? Round all dollar values to the nearest cent.
Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0
The question is an annuity question with the present value of the annuity given.
The present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / (r/t) where PV = $61,600; r = interest rate = 9.84% = 0.0984; t = number of payments in a year = 6; n = number of years = 11 years and P is the periodic payment.
61600 = P(1 - (1 + 0.0984/6)^-(11 x 6)) / (0.0984 / 6)
61600 = P(1 - (1 + 0.0164)^-66) / 0.0164
61600 x 0.0164 = P(1 - (1.0164)^-66)
1010.24 = P(1 - 0.341769) = 0.658231P
P = 1010.24 / 0.658231 = 1534.78
Thus, Niki pays $1,534.78 every two months for eleven years.
The total payment made by Niki = 11 x 6 x 1,534.78 = $101,295.48
Therefore, interest paid by Niki = $101,295.48 - $61,600 = $39,695.48
malfutka [58]3 years ago
4 0

Answer:

answer is A on edge

Step-by-step explanation:

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Which number is a multiple of 7
harina [27]

Answer:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70...

Step-by-step explanation:

6 0
3 years ago
If the filling equipment is functioning properly what is the probability that the volume of oil in a randomly selected barrel wi
Sophie [7]

Answer:

P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)

P(Z>0.8)=1-P(Z\leq 0.8)

P(Z>0.8)=1-0.788=0.212

Step-by-step explanation:

Assuming this previous info : "Trucks carry barrels of crude oil from a port in Texas to a distributer in New Mexico on long  trailers. Filling equipment is used to fill the barrels with the oil. When functioning properly,  the actual volume of oil loaded into each barrel by the filling equipment at the port is  approximately normally distributed with a mean of 55 gallons and a standard deviation of 0.5  gallons. If the mean is greater than 55.4 gallons, the filling mechanism is overfilling."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the amount filling of a population, and for this case we know the distribution for X is given by:

X \sim N(55,0.5)  

Where \mu=55 and \sigma=0.5

We are interested on this probability

P(X>55.4)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)

And we can find this probability using the complement rule:

P(Z>0.8)=1-P(Z\leq 0.8)

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(Z>0.8)=1-0.788=0.212

8 0
3 years ago
The product of x and the sum of 6 and 8 times the square of x
Vaselesa [24]

Answer:

x(6 + 8x²) or 6x + 8x³.

Step-by-step explanation:

"The square of x" can be represented by x² and 8 times that would be 8 * x² or 8x². The sum of 6 and 8x² can be represented by 6 + x² and the product of x and 6 + x² can be represented by x * (6 + 8x²) or x(6 + 8x²) which simplifies to 6x + 8x³.

7 0
3 years ago
Read 2 more answers
A child lets go of a balloon that rises at a constant rate. 5 seconds after it was released, the balloon is at a height of 16 fe
Mama L [17]

Answer:

1. h = 2.4t + 4

2. 4 feet

3. 220 feet

Step-by-step explanation:

1. Write a linear model for the height, h, of the balloon as a function of the number of seconds, s that it has been raising.

Since the balloon rises at a constant rate, we find this rate by using the initial and final values of height and time at 5 seconds and 20 seconds respectively which are 12 feet and 52 feet respectively.

So, rate = gradient of line

= change in height/change  in time

= (52 ft - 16 ft)/(20 s - 5 s)

= 36 ft/15 s

= 2.4 ft/s

Now the equation of the line which shows the height is gotten from

(h - h')/(t - t') = rate

Using h'= 16 feet and t' = 5 s, we have

(h - 16)/(t - 5) = 2.4

h - 16 = 2.4(t - 5)

h - 16 = 2.4t - 12

h = 2.4t - 12 + 16

h = 2.4t + 4

where h is the height of the balloon above the ground and t is the time spent in the air in seconds.

2. What was the height of the balloon initially before the child let it go?

We obtain the initial height of the balloon before the child let go at time, t = 0 the time before the child let go.

So, substituting t = 0 into the equation for h, we have

h = 2.4t + 4

h = 2.4(0) + 4

h = 0 + 4

h = 4 feet

So, the height of the balloon before the child let go is 4 feet above the ground.

3. Use your model to predict the height of the balloon after 90 seconds.

We insert t = 90 s into the equation for h. So,

h = 2.4t + 4

h = 2.4(90) + 4

h = 216 + 4

h = 220 feet

So, the height of the balloon after 90 s is 220 feet above the ground.

8 0
2 years ago
Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicin
aliya0001 [1]

Answer:

Step-by-step explanation:

Consider the graphs of the y = 4x  and  y = \frac{x^{2} }{4}.

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

4x = \frac{x^{2} }{4} \\   x^{2}  = 16x \\ x^{2}  - 16x = 0 \\   x(x-16) = 0 then x=0  o  x=16. Therefore the integration limits are:

y = 4(0) = 0  and  y = 4(16) = 64

The inverse functions are given by:

x = 2 \sqrt{y}  and  x = \frac{y}{4}. Then

The volume of the solid of revolution is given by:

\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2}  dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\  dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}

6 0
3 years ago
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