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Tpy6a [65]
3 years ago
8

Sam promises to pay Sandy $2,000 in four years and another $3,000 four years later for a loan of $2,000 from Sandy today. What i

s the interest rate that Sandy is getting? Assume interest is compounded monthly. A. 14.75% B. 16.72% C. 15.10% D. 18.08%
Engineering
1 answer:
frozen [14]3 years ago
8 0

Answer:

the interest rate that Sandy is getting is (C.) 15.10%

Explanation:

Given data

in 4 year cash pay (p1)  = $2000

in 8 year cash pay (p2)  = $3000

time period (t1) = 4 years

time period (t2) =  8 years

loan value = $2000

To find out

interest rate

solution

first we know amount will be paid in first 4 year is

$2000 (1+r/100)^{12t}

$2000 (1+r/100)^{48}                  ...................1

now we calculate the next payment will paid after 4 year i.e.

$2000 (1+r/100)^{12t} - $2000  

$2000 (1+r/100)^{48}   - $2000  ..................2

after full time period of payment total amount will be paid by equation 1 and 2  i.e.

$3000 = $2000 (1+r/100)^{48} ×$2000 (1+r/100)^{48} - $2000

$3000 = $2000 ( (1+r/100)^{48} ×  (1+r/100)^{48} - 1 )   .....3

now we have solve   (1+r/100)^{48} this eqution

so we consider  (1+r/100)^{48} = A

so new equation will be by equation 3

$3000/$2000 = ( A - 1 ) × A

3/2 = A² - A

solve this equation we get  2A² - 2 A - 3 = 0 so A = 1.823

now we compute A again in  (1+r/100)^{48} = A

(1+r/100)^{48} = 1.823

so rate (r) = 1.258 % / month

and rate yearly = 1.258 ×12

the interest rate that Sandy is getting yearly 15.10 %

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Answer:

Vf = specific volume of saturated liquid  = 0.0217158 ft^3/lb

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Explanation:

Given data:

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Vf = specific volume of saturated liquid  

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we know that from the saturated steam table we can find these value

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In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
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Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

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The probatility function is given by :

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Where nCx is the combinatorial number

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Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

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