To solve this we are going to use the simple interest formula:
[tex]A=P(1+rt)[/tex]
where
[tex]A[/tex] is the final amount after [tex]t[/tex] years
[tex]P[/tex] is the initial amount
[tex]r[/tex] is the interest rate in decimal form
[tex]t[/tex] is the time in years
A. We now form our problem that her initial savings were $4000. Since she earned $960 in interest after 3 years, her final amount will be his initial savings plus the interest hat she earned:
[tex]A=4000+960[/tex]
[tex]A=4960[/tex]
We can conclude that she will have $4960 in her account at the end of three years.
B. We know for our problem that [tex]P=4000[/tex], and [tex]t=3[/tex]. We also know that the account earned $960 in interest, so [tex]A=4000+960=4960[/tex]. Let [tex]x[/tex] represent the interest rate; since the interest rate should be in decimal form, [tex]r= \frac{x}{100} =0.01x[/tex]. Lets replace all the values in our formula to find [tex]x[/tex]:
[tex]4960=4000[1+(0.01x)(3)][/tex]
[tex]4960=4000(1+0.03x)[/tex]
[tex]4960=4000+120x[/tex]
[tex]120x=960[/tex]
[tex]x= \frac{960}{120} [/tex]
[tex]x=8[/tex]
We can conclude that the account growth by an annual simple interest rate of 8%
C. if the interest rate were 1% greater, our new [tex]r[/tex] is going to be [tex]r=8+9[/tex]. But remember that the interest rate should by in decimal form, so we are going to divide 9% by 100%:
[tex]r= \frac{9}{100} [/tex]
[tex]r=0.09[/tex]
We also know that the conditions are the same as before, so [tex]P=4000[/tex] and [tex]t=3[/tex]. Lets replace all our values in our formula to find [tex]A[/tex]:
[tex]A=4000[1+(0.09)(3)][/tex]
[tex]A=4000(1+0.27)[/tex]
[tex]A=4000(1.27)[/tex]
[tex]A=5080[/tex]
We can conclude that she will have [tex]5080-4960=120[/tex] $ more in her account if the interest rate were 1% greater.
