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2 years ago

the length of the highway is 1500 miles. If 1 inch represents 150 miles, what is the length of the highway on the map?

Mathematics

1 answer:

NeX [460]2 years ago

4 0

Answer:

10 inches.

Step-by-step explanation:

If 1 inch represents 150 miles, and there are 1500 miles covered by the map, then the length of the highway on the map is 10 inches.

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1. You are saving to buy a new house in 7 years. If you invest $4,500 now at 5.5% interest compounded

motikmotik

Answer:

Part 1) $\$6,595.94$

Part 2) $\$3,449.23$

Part 3) $\$17,040.06$

Part 4) $\$20,773.90$

Part 5) The Option A is the best way to invest the money by $4,223.94 than Option B

Step-by-step explanation:

Part 1)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=7\ years\\ P=\$4,500\\ r=5.5\%=5.5/100=0.055\\n=4$

substitute in the formula above

$A=4,500(1+\frac{0.055}{4})^{4*7}$

$A=4,500(1.01375)^{28}$

$A=\$6,595.94$

Part 2)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=2\ years\\ P=\$3,200\\ r=3.75\%=3.75/100=0.0375$

substitute in the formula above

$A=3,200(e)^{0.0375*2}$

$A=\$3,449.23$

Part 3)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=18\ years\\ A=\$40,000\\ r=4.75\%=4.75/100=0.0475\\n=12$

substitute in the formula above

$40,000=P(1+\frac{0.0475}{12})^{12*18}$

$40,000=P(\frac{12.0475}{12})^{216}$

$P=40,000/[(\frac{12.0475}{12})^{216}]$

$P=\$17,040.06$

Part 4)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=7\ years\\ A=\$30,000\\ r=5.25\%=5.25/100=0.0525$

substitute in the formula above

$30,000=P(e)^{0.0525*7}$

$30,000=P(e)^{0.3675}$

$P=30,000/(e)^{0.3675}$

$P=\$20,773.90$

Part 5)

Option A

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=8\ years\\ P=\$11,500\\ r=5.6\%=5.6/100=0.056\\n=2$

substitute in the formula above

$A=11,500(1+\frac{0.056}{2})^{2*8}$

$A=11,500(1.028)^{16}$

$A=\$17,889.07$

Option B

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=5\ years\\ P=\$11,500\\ r=3.45\%=3.45/100=0.0345$

substitute in the formula above

$A=11,500(e)^{0.0345*5}$

$A=11,500(e)^{0.1725}$

$A=\$13,665.13$

Compare the options

Option A ------> $\$17,889.07$

Option B -----> $\$13,665.13$

Option A > Option B

Find out the difference

$\$17,889.07-$13,665.13=$4,223.94$

therefore

The Option A is the best way to invest the money by $4,223.94 than Option B

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