ddaeeric
2020-12-02
Answered

If $2000 is invested at an interest rate of 4.5% per year, compounded continuously, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)
a) 2 years
b) 4 years
c) 12 years

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jlo2niT

Answered 2020-12-03
Author has **96** answers

Analyse the given information
Since the interest is compounded continuously, the Principal will increase exponentially.
The amount after t years is given by $A\left(t\right)=P{e}^{rt}$
It is given that the initial investment,$P=\mathrm{\$}2000}$
and the interest rate per year,$r=4.5\mathrm{\%}=0.045$
Process:
In each part a), b) and c) plug in values of P, r and t in each case to find the value of Amount after a given number of years.
To round to nearest cent we need to round answers to two decimals.
a) After two years i.e t=2
$A\left(t\right)=P{e}^{rt}$
After two years: t=2
$A\left(2\right)=2000{e}^{0.045\left(2\right)}$

$A\left(2\right)=2000{e}^{0.09}$

$A\left(2\right)=\mathrm{\$}2188.35}$ [rounded to two decimals(nearest cent) ]
Amount after 2 years is $2188.35
b) After two years i.e t=4
$A\left(t\right)=P{e}^{rt}$
After two years: t=4
$A\left(4\right)=2000{e}^{0.045\left(4\right)}$

$A\left(4\right)=2000{e}^{0.18}$

$A\left(4\right)=\mathrm{\$}2394.43}$ [rounded to two decimals(nearest cent) ]
Amount after 4 years is $2394.43
c) After twelve years i.e t=12
$A\left(t\right)=P{e}^{rt}$
After two years: t=12
$A\left(12\right)=2000{e}^{0.045\left(12\right)}$

$A\left(12\right)=2000{e}^{0.54}$

$A\left(12\right)=\mathrm{\$}3432.01}$ [rounded to two decimals(nearest cent)]
Amount after 12 years is $3432.01

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Let (X,M,$\mu $) be a measure space and $f:X\to [0,\mathrm{\infty}]$ a measurable function. Prove that

${\int}_{X}fd\mu >0\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\mu {\textstyle (}\{x\in X\mid f(x)>0\}{\textstyle )}>0.$

I tried to use the definition of the integral by the approximation theorem of measurable functions by simple functions, but I haven't been able to come to the conclusion.

Let (X,M,$\mu $) be a measure space and $f:X\to [0,\mathrm{\infty}]$ a measurable function. Prove that

${\int}_{X}fd\mu >0\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\mu {\textstyle (}\{x\in X\mid f(x)>0\}{\textstyle )}>0.$

I tried to use the definition of the integral by the approximation theorem of measurable functions by simple functions, but I haven't been able to come to the conclusion.