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Sagot :
initial investment, P = $1000
Linconln Federal Bank- LFB interest rate, 6% compounded yearly
Whashington National Bank- WNB interest rate, 5.5% compounded daily
a. exponential function
We'll use the compound interest formula
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where,
A=final amount
P=initial principal balance
r=interest rate
n=number of times interest applied per time period
t=number of time periods elapsed
we know that
P = $1000
t is in years
LFB, r = 0.06 compounded yearly , then n = 1
[tex]\begin{gathered} A_{LFB}=1000(1+0.06)^t \\ A_{LFB}=1000*1.06^t \end{gathered}[/tex]WNB, r = 0.05 compounded daily
[tex]A_{WNB}=1000(1+\frac{0.055}{365})^{365*t}[/tex]b
Let' see, if t = 10 years, then A-LFB = $1790.85 and A-WNB = $1733.18
Caleb should choose LFB
c.
the domain of these exponential functions is
[tex]D:(-\infty,\infty)[/tex]the range is,
[tex]R:(0,\infty)[/tex]This functions have a horizontal asymptote at y = 0
behavior, since the base of the exponent is >1 this is an increasing function
intercepts,
[tex]\begin{gathered} x-axis:NONE \\ y-axis:(0,1000) \end{gathered}[/tex]
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