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DIA [1.3K]
3 years ago
14

Using technology, determine the monthly payment on a 6 year loan of $15,250 at 3.5% compounded monthly. Round your answer to the

nearest cent.
a. $234.75c. $582.70b. $235.13d. $590.05
Mathematics
2 answers:
8_murik_8 [283]3 years ago
4 0
235.13 hope it helps
Gekata [30.6K]3 years ago
3 0

Answer:

Option 3 or b - The monthly payment is $235.13.                                                                        

Step-by-step explanation:

Given : A 6 year loan of $15,250 at 3.5% compounded monthly.  

To find :  Monthly payment ?

Solution :

The formula to find monthly payment is

M=\frac{\text{Amount}}{\text{Discount factor}}

Discount factor is D=\frac{1-(1+i)^{-n}}{i}

Substitute in the formula,

M=\frac{A}{\frac{1-(1+i)^{-n}}{i}}

M=\frac{A\times i}{1-(1+i)^{-n}}

where, A is the amount A=$15250

r is the rate = 3.5%=0.035 compounded monthly

i=\frac{r}{12}=\farc{0.035}{12}=0.00291

time t=6 years

Time in months n=t\times 12=6\times 12=72

Substitute all the values in the formula,

M=\frac{15250\times 0.00291}{1-(1+0.00291)^{-72}}

M=\frac{44.479}{1-(1.00291)^{-72}}

M=\frac{44.479}{1-0.8112}

M=\frac{44.479}{0.18916}

M=235.13

Therefore, The monthly payment is $235.13.

So, Option 3 or b is correct.

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Confidence interval can be calculated by below formula

Pcap - Z_{\alpha /2}\sqrt{\frac{Pcap(1 - Pcap)}{n} } < p < Pcap + Z_{\alpha /2}\sqrt{\frac{Pcap(1 - Pcap)}{n} }

where the left part is the lower interval and right side is the higher interval.

Pcap = 750/1100

Pcap = 0.6818

Again to calculate Z_{\alpha /2} , as we know that the level of confidence is 90 percent so its value will be 1.645

now putting the values on the both side we can get ,

0.6818 - 1.645 x 0.014 < p  <  0.6818 + 1.645 x 0.014

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the nearest value is [0.6592, 0.7044]

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5 0
1 year ago
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Answer:

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Students at a certain school were​ surveyed, and it was estimated that 10​% of college students abstain from drinking alcohol. T
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Answer:

a)

<em> you are unwilling to predict the proportion value at your school  = 0.90</em>

<em>b) </em>

<em>The large sample size 'n' = 864</em>

<em></em>

Step-by-step explanation:

Given estimated proportion 'p' = 10% = 0.10

<em>Given Margin of error M.E = 0.02</em>

<em>Level of significance α = 0.05</em>

a)

<em>  you are unwilling to predict the proportion value at your school </em>

<em>   q = 1- p = 1- 0.10 =0.90</em>

b)

<em>The Margin of error is determined by</em>

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Cross multiplication , we get

                      \sqrt{n}  = \frac{1.96 X \sqrt{0.10 (0.90)} }{0.02 }

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<em>Squaring on both sides , we get</em>

<em>                      n = 864.36</em>

<u><em>Conclusion</em></u><em>:-</em>

<em>The large sample size 'n' = 864</em>

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