Question

A savings account compounds interest, at a rate of 17%, once a year. John puts $1,000 in the account as the principal. How can John set up a function to track the amount of money he has?
A(x) = 1000(17)x where 17 is the interest rate
A(x) = 1000(.17)x where .17 is the interest rate
A(x) = 1000(1 + .17)x where .17 is the interest rate
A(x) = 1000(1 + 17)x where 17 is the interest rate

Answer 1

Answer:

C. $A(x)=1000\cdot (1+0.17)^x$, where 0.17 is the interest rate.

Step-by-step explanation:

We have been given that a savings account compounds interest, at a rate of 17%, once a year. John puts $1,000 in the account as the principal. We are asked to set up a function to track the amount of money John has.

We will use exponential form of function to set up required function.

An exponential function is in form $y=a\cdot b^x$, where,

a = Initial value,

b = For growth b is in form (1+r), where r represents the growth rate in decimal form.

Let us convert our given rate in decimal form.

$17\%=\frac{17}{100}=0.17$

Since John invested an amount of $1,000, so a=1000 for our growth function.

Upon substituting our given values in growth function we will get,

$A(x)=1000\cdot (1+0.17)^x$, where A(x) represents amount of money after x years.

Therefore, $A(x)=1000\cdot (1+0.17)^x$ is our required function and option C is the correct choice.

Answer 2

#3
A(x) = 1000(1+.17)x

because the 17 percent would have to be changed to a decimal form