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Talja [164]
4 years ago
15

Melanie is looking for a loan. She is willing to pay no more than an effective rate of 9.955% annually. Which, if any, of the fo

llowing loans meet Melanie’s criteria? Loan A: 9.265% nominal rate, compounded weekly Loan B: 9.442% nominal rate, compounded monthly Loan C: 9.719% nominal rate, compounded quarterly
Mathematics
2 answers:
Y_Kistochka [10]4 years ago
5 0

Answer:

Loan A and B

Step-by-step explanation:

Since, the effective annual interest rate is,

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

Where, r is the nominal rate( in decimals) per period,

n is the number of periods,

In Loan A :

r = 9.265% = 0.09265,

n = 52,

Thus, the effective annual interest rate is,

i_1=(1+\frac{0.09265}{52})^{52}-1

=(1+0.00178173077)^{52}-1

=1.09698725072-1

=0.09698725072\approx 0.9670

\implies i_1=9.670\%

In Loan B :

r = 9.442% = 0.09442,

n = 12,

Thus, the effective annual interest rate is,

i_2=(1+\frac{0.09442}{12})^{12}-1

=(1+0.00786833)^{12}-1

=1.09861519498-1

=0.09861519498\approx 0.9862

\implies i_2=9.862\%

In Loan C :

r =9.719% = 0.09719,

n = 4,

Thus, the effective annual interest rate is,

i_3=(1+\frac{0.09719}{4})^{4}-1

=(1+0.0242975)^{4}-1

=1.10078993749-1

=0.10078993749\approx 0.10079

\implies i_3=10.079\%

Since, i_1, i_2 but i_3>9.955\%

Hence, Loan A and B meets his criteria.

Bezzdna [24]4 years ago
4 0
The answers are both loan A and loan B :)

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amm1812

Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

       \sigma  =  \sqrt{\frac{ p(1- p)}{n} }

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

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So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

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