Mathematics, 26.02.2020 06:31, skullglitches

Norah has $50,000 to invest. She is considering two investment options. Option A pays 1.5% simple interest. Option B pays 1.4% interest compounded annually. Drag dollar amounts to the table to show the value of each investment option after 5 years, 10 years, and 20 years rounded to the nearest dollar.

Answers: 2

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Answers

answered: Turtlelover05

Part 1) Option A

a) $A=\$53,750$

b) $A=\$57,500$

c) $A=\$65,000$

Part 2) Option B

a) $A=\$53,864$

b) $A=\$58,027$

c) $A=\$67,343$

Step-by-step explanation:

Part 1) Option A

Simple Interest

we know that

The simple interest formula is equal to

A=P(1+rt)

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

case a) 5 years

$t=5\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015$

substitute in the formula above

A=50,000(1+0.015*5)

A=50,000(1.075)

$A=\$53,750$

case b) 10 years

$t=10\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015$

substitute in the formula above

A=50,000(1+0.015*10)

A=50,000(1.15)

$A=\$57,500$

case c) 20 years

$t=20\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015$

substitute in the formula above

A=50,000(1+0.015*20)

A=50,000(1.30)

$A=\$65,000$

Part 2) Option B

interest compounded annually

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

case a) 5 years

$t=5\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1$

substitute in the formula above

$A=50,000(1+\frac{0.015}{1})^{1*5}$

$A=50,000(1.015)^{5}$

$A=\$53,864$

case b) 10 years

$t=10\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1$

substitute in the formula above

$A=50,000(1+\frac{0.015}{1})^{1*10}$

$A=50,000(1.015)^{10}$

$A=\$58,027$

case c) 20 years

$t=20\ years\\ P=\$50,000\\r=1.5\%=1.5/100=0.015\\n=1$

substitute in the formula above

$A=50,000(1+\frac{0.015}{1})^{1*20}$

$A=50,000(1.015)^{20}$

$A=\$67,343$

answered: Guest

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

when multiplying or dividing by a negative on both sides you have to flip the inequality sign. this rule only applies to multiplying or dividing by a negative number in inequality equations. this rule is really important.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice

5c+2< 27 i really do not understand this.

answered: Guest

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2 is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes. our only condition is that it passes through the point (-2,5)

A. write an equation of a line that passes through (-2, 5). b. how many different equations could y

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Norah has $50,000 to invest. She is considering two investment options. Option A pays 1.5% simple in...