m the account, muzic-ivan.info the adhering to questions.A. just how much will certainly be in the account after three years?B.How much will be in the account after 18 years? C. How many type of years will certainly it take for the account to contain $2500? D. How many type of years will certainly it take for the account to contain $3000?

The formula for compound growth is:

Where,

P is the future value is the initial despositr is the yearly price of interestn is the number of times compounding happens ( n = 1 once annual compounding, n = 2 when semi-yearly compounding, n = 4 as soon as quarterly compounding etc.)t is time in years

, r is 0.04, n is 1 (considering that annual compounding) and also t is 3. dollars

B.

Everypoint is exact same in this problem as component A, simply t is not 3, t is 18. Similarly, we solve:

dollars

C.

Here we desire to number out t, provided P=2500,

, r is 0.04, n is 1 (considering that annual compounding). We have:

To fix this exponential component, we take Natural Logarithm (ln) and usage our properties of logarithms and fix for t.

\ln(1.25)=t*ln(1.04)\t=fracln(1.25)ln(1.04)\t=5.69" alt="ln(1.25)=ln<1.04^t>\ln(1.25)=t*ln(1.04)\t=fracln(1.25)ln(1.04)\t=5.69" align="absmiddle" class="latex-formula"> years

It is equivalent to component C, just P=3000 instead of P=2500. Let us setup the equation and also fix for t.

\ln(frac32)=t*ln(1.04)\t=fracln(frac32)ln(1.04)\t=10.34" alt="3000=2000(1+0.04)^t\frac30002000=(1.04)^t\frac32=(1.04)^t\ln(frac32)=ln<1.04^t>\ln(frac32)=t*ln(1.04)\t=fracln(frac32)ln(1.04)\t=10.34" align="absmiddle" class="latex-formula"> years

muzic-ivan.info:

A. $2249.73

B. $4051.63

C. 5.69 years

D. 10.34 years

Send

liraira <26>9 months ago 5 0 Hi thereThe formula isA=p (1+r)^tA future valueP existing valueR interemainder rateT timeA) A=2,000×(1+0.04)^(3)=2,249.728B) A=2,000×(1+0.04)^(18)=4,051.63C) 2500=2000 (1+0.04)^tSolve for tT=log(2,500÷2,000)÷log(1+0.04)T=5.7 yearsD) t=log(3,000÷2,000)÷log(1+0.04)t=10.3 yearsHope it helps Sfinish You might be interested in

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