Solve. Assume the exercise describes a linear relationship. When writing a linear equation, write the equation in
slope-intercept form.
-In 1985, John invested $15,000 in the stock market. By 1994 his investment had grown to $16,300. If the market continues to grow at the same rate, how much will be in his account in 1996? Give your answer to the nearest
Dollar.
A)$17,600
B)$16,300
C)$16,589
D)$15,290

Question

Quizplus · Accepted Answer

We can assume a linear relationship between the amount invested and the worth of the investment. We can find the rate of growth using the initial and final amounts invested over the given time period:

Rate of growth = (final amount - initial amount) / initial amount x 100%
Rate of growth = (16,300 - 15,000) / 15,000 x 100%
Rate of growth = 8.67%

Using this rate of growth, we can predict the worth of John's investment in 1996, two years after 1994:

Investment in 1996 = $16,300 x (1 + 0.0867)^2
Investment in 1996 = $16,300 x 1.186511
Investment in 1996 ≈ $16,589

Therefore, the best choice is C, $16,589. Let x be the number of years since 1985. 
The initial investment, y-intercept, is $15,000. 
The final amount, after 9 years (1994), is $16,300.
Using these two points, we can find the slope-intercept form of the linear equation:
y = mx + b
where m is the slope (rate of growth) and b is the y-intercept.

m = (y2 - y1)/(x2 - x1) = ($16,300 - $15,000)/(9 - 0) ≈ $156.67/year
b = $15,000

Thus, the equation is: y = $156.67x + $15,000

To find out how much John's investment will be in 1996, we need to plug in x = 11 (since 1996 is two years after 1994) and solve for y:

y = $156.67(11) + $15,000 ≈ $16,589

Therefore, the closest answer choice is C) $16,589.

Solve Assume the Exercise Describes a Linear Relationship