10.How much would $1,000 in an account paying 14 percent interest compounded sem
ID: 2819803 • Letter: 1
Question
10.How much would $1,000 in an account paying 14 percent interest compounded semi-annually accumulate to in 10 years? a. $2,140 b. $3,707 c. $1,647 d. $3,870 11.lf you want to have $90 in four years, how much money must you put in a savings account today? Assume that the savings account pays 8.5% and it is compounded monthly (round to the nearest a. $64 b. $65 c. $66 d. $71 12.What is the present value of $12,500 to be received 10 years from today? Assume a discount rate of 8% compounded annually and round to the nearest $10 a. $5,790 b. $11,574 c. $9,210 d. $17,010 13.If you want to have $1,200 in 27 months, how much money must you put in a savings account today? Assume that the savings account pays 14% and it is compounded monthly (round to nearest $10) a. $910 b. $890 c. $880 d. $860 14.If you want to have $2,100 in 3 years, how much money must you put in a savings account today? Assume that the savings account pays 7% and it is compounded quarterly a. $1,656 b. $1,705 c. $1,674 d. $1,697Explanation / Answer
10.
We use the formula:
A=P(1+r/200)^2n
where
A=future value
P=present value
r=rate of interest
n=time period.
A=$1000(1+0.14/2)(2*10)
=$1000*3.869684462
which is equal to
=$3870(Approx).
11.
We use the formula:
A=P(1+r/1200)^12n
where
A=future value
P=present value
r=rate of interest
n=time period.
90=P(1+0.085/12)^(12*4)
P=90/(1+0.085/12)^(12*4)
=(90*0.712623898)
which is equal to =$64(Approx).
12.
We use the formula:
A=P(1+r/100)^n
where
A=future value
P=present value
r=rate of interest
n=time period.
12500=P(1.08)^10
P=12500/1.08^10
=(12500*0.463193488)
which is equal to
=$5790(Approx).
13.
We use the formula:
A=P(1+r/1200)^12n
where
A=future value
P=present value
r=rate of interest
n=time period.
1200=P(1+0.14/12)^27
P=1200/(1+0.14/12)^27
=1200*0.731120735
which is equal to
=$880(Approx).
14.
We use the formula:
A=P(1+r/400)^4n
where
A=future value
P=present value
r=rate of interest
n=time period.
2100=P(1+0.07/4)^(4*3)
P=2100/(1+0.07/4)^(4*3)
=2100*0.81205788
which is equal to
=$1705(Approx).
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