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Pepsi [2]
3 years ago
6

Find three consecutive even integers whose sum is 72

Mathematics
2 answers:
Reil [10]3 years ago
7 0

Answer:

23, 24 , 25

Step-by-step explanation:

23+24+25=72

photoshop1234 [79]3 years ago
4 0
23,24,25 is the answer
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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)

<u><em>Option A</em></u>

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  

<u><em>Option B</em></u>

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  

so

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

3 0
3 years ago
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