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Nezavi [6.7K]
3 years ago
14

In Exercise,find the balance in the account after 10 years.

Mathematics
1 answer:
djverab [1.8K]3 years ago
3 0

Answer: The balance in the account after 10 years is $3374.65

Step-by-step explanation:

The exponential equation for growth [ compounded continuously] is

y=pe^{rt}

, where P= Present value

r= growth rate  ( in decimal)

t= time (years)

By considering the given information , we have

p=$2500  r = % =0.03  and t= 10

Substitute all the values in the above equation , we get

y=2500e^{0.03(10)}

y=2500e^{0.3}

y=2500(1.3498588)=3374.64701894\approx3374.65  [Round to the nearest cent]

Therefore, the balance in the account after 10 years is $3374.65

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A company that rents small moving trucks wants to purchase 25 trucks with a combined capacity of 28,000 cubic feet. Three differ
pickupchik [31]

Answer:

We have 4 solutions:

  • No 10-foot truck, 10  14-foot trucks, and 15 24-foot trucks
  • 2 10-foot trucks, 7 14-foot trucks, and 16 24-foot trucks
  • 4 10-foot trucks, 4  14-foot trucks, and 17 24-foot trucks
  • 6 10-foot trucks, 1 14-foot trucks, and 18 24-foot trucks

Step-by-step explanation:

Let the number of 10-foot truck with a capacity of 350 cubic feet purchased=a

Let the number of 14-foot truck with a capacity of 700 cubic feet purchased=b

Let the number of 24-foot truck with a capacity of 1,400 cubic feet purchased=c

The company wants to purchase 25 trucks, therefore.

  • a+b+c=25

Furthermore, the combined capacity of the trucks is 28,000 cubic feet.

  • 350a+700b+1400c=28000

Since the number of equations is less than the number of variables, you can not use a matrix equation to solve this problem.  The solution is most easily found using an augmented matrix.  

The augmented matrix is presented below:  

\left[\begin{array}{ccc|c}1&1&1&25\\350&700&1400&28000\end{array}\right]

Using the calculator, the reduced row echelon form is:

\left[\begin{array}{ccc|c}1&0&-2&-30\\0&1&3&55\end{array}\right]

where  

a- 2c=-30 means a =2c-30

b+3c=55 means b= 55-3c

We alter the value of c as long as neither a nor b becomes negative. Suitable values for c are 15, 16, 17, and 18:

\left|\begin{array}{|c||c||c|}a=2c-30&b=55-3c&c\\0&10&15\\2&7&16\\4&4&17\\6&1&18\end{array}\right|

We can easily  verify that, for each solution, the number of trucks adds up to 25 and the fleet capacity is 28,000 cubic feet.

We therefore have 4 solutions:

  • No 10-foot truck, 10  14-foot trucks, and 15 24-foot trucks
  • 2 10-foot trucks, 7 14-foot trucks, and 16 24-foot trucks
  • 4 10-foot trucks, 4  14-foot trucks, and 17 24-foot trucks
  • 6 10-foot trucks, 1 14-foot trucks, and 18 24-foot trucks
4 0
3 years ago
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