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Tpy6a [65]
2 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]2 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:

Explanation along with code and output results is provided below.

C++ Code:

#include <iostream>

using namespace std;

int main()

{

   int year;

   cout<<"Enter the car model year."<<endl;

   cin>>year;    

  if (year<=1969)

  {

cout<<"Few safety features."<<endl;

  }

else if (year>=1970 && year<1989)

{

cout<<"Probably has seat belts."<<endl;

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cout<<"Probably has antilock brakes."<<endl;

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cout<<"Probably has airbags."<<endl;

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   return 0;

}

Explanation:

The problem was to print feature messages of a car given its model year.

If else conditions are being used incorporate the logic. The code has been tested with several inputs and got correct output results.

Output:

Enter the car model year.

1961

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Enter the car model year.

1975

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1994

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5 0
3 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 3 m. If the height of the tank sides is 4 m. What
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Answer:

ay max = 4.91 m/s²  

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Explanation:

given data

wide b = 1 m

long l  = 2 m

depth d = 3 m

height of  tank sides h = 4 m

solution

here for prevent spilling condition is

\frac{dz}{dy} ≤ - \frac{1.5 - 1 }{1}  ..........1

\frac{dz}{dy}  ≤ - 0.50

and when here

\frac{dz}{dy} =  -   \frac{ay}{g + az}    ......2  

when az is 0 ay will be

ay = -  \frac{dz}{dy} g    

and ay max will be

ay max = -( -0.50) (9.81 )

ay max = 4.91 m/s²  

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8 0
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