What is the value of a $2500 investment after it has grown by 6.3% for 10 years?

The value of the investment would be $4,075

Step-by-step explanation: The initial investment, that is, the principal is given as $2500, and it was invested for a period of 10 years at the rate of 6.3%. Using the simple interest calculation formula, the investment would have a present value as follows;

Interest = P x R x T

Where P = amount invested (2500), R = rate of interest (6.3 or 0.063) and T = number of years (10)

Interest = 2500 x 0.063 x 10

Interest = 1575

The value of the investment at 6.3% after 10 years is now;

Initial investment + Interest

2500 + 1575 = 4075

Therefore the value of the investment is $4,075

Step-by-step explanation:

we know that

The linear equation in slope intercept form is equal to

where

m is the slope

b is the y-intercept

In this problem

Let

y is the cost of a telephone call

x is the number of minutes

The slope is equal to

The y-intercept is equal to

substitute

For x=T minutes

substitute

therefore

The expression that models the cost of a telephone call that last T minutes is