Mathematics, 26.06.2019 10:10, ashiteru123

What will be the simple interest earned when you invest $1,000 for 3 years at 10 percent and the compound interest earned when you invest the same sum for 2 years at 5 percent ? the simple interest earned when you invest $1,000 for 3 years at 10 % is $ . the interest compounded when you invest the same sum for 2 years at 5 % is $ . there are 7 trout fish in a pond, and the population doubles every year. find the population after t years. arrowboth a company buys a machine for $3,000. the value of the machine depreciates by 7% every year. find the value of the machine after t years. arrowboth the initial population of a colony of ants is 300. the number of ants increases at a rate of 1.5% every month. find the population of ants after t months. arrowboth a research laboratory is testing a new vaccine on 300 infected cells. the decay rate is 1.5% per minute. find the number of infected cells after t minutes. arrowboth

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Answers

answered: johanperezpere51

Part 1) The simple interest earned when you invest $1,000 for 3 years at 10 % is $300

Part 2) The interest compounded when you invest the same sum for 2 years at 5 % is $102.50

Part 3) f(t)=7(2)^t

Part 4) f(t)=3,000(0.93)^t

Part 5) f(t)=300(1.015)^t

Part 6) f(t)=300(0.985)^t

Step-by-step explanation:

Part 1) What will be the simple interest earned when you invest $1,000 for 3 years at 10 percent

we know that

The simple interest formula is equal to

I=P(rt)

where

I is the Final Interest 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

$t=3\ years\\ P=\$1,000\\r=0.10$

substitute in the formula above

$I=\$1,000(0.10*3)=\$300$

Part 2) What will be the compound interest earned when you invest $1,000 for 2 years at 5 percent ?

we know that

The compound interest formula is equal to

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

where

I 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

$t=2\ years\\ P=\$1,000\\ r=0.05\\n=1$

substitute in the formula above

$A=\$1,000(1+\frac{0.05}{1})^{1*2}$

$A=\$1,000(1.05)^{2}=\$1,102.50$

The interest is equal to

$I=\$1,102.50-\$1,000=\$102.50$

Part 3) There are 7 trout fish in a pond, and the population doubles every year.

Find the population after t years.

we know that

This question is about exponent function of the form

f(x)= a(b)^t

where

a is the initial value

b is the base of the exponent.

In this problem we have

There are 7 trout fish in the pound ----> initial value a=7

The population is double every year ------> the base is b=2

substitute

f(t)= 7(2)^t

Part 4) A company buys a machine for $3,000. The value of the machine depreciates by 7% every year. Find the value of the machine after t years.

we know that

This question is about exponent function of the form

f(x)= a(b)^t

where

a is the initial value

b is the base of the exponent.

we have

Company buys a machine for $3,000 --> initial value is a=3,000

The value depreciate 7% a year

Since it was decreased by 7% every year, it will become: 100%-7%=93%

the base is 93%, b=0.93

substitute

f(t)=3,000(0.93)^t

Part 5) The initial population of a colony of ants is 300. The number of ants increases at a rate of 1.5% every month. Find the population of ants after t months.

we know that

This question is about exponent function of the form

f(x)= a(b)^t

where

a is the initial value

b is the base of the exponent.

we have

Initial population of ants is 300----> initial value is a=300

The number of ants increases 1.5% per month.

Since it will increases by 1.5% every month, it will become: 100%+1.5%=101.5%

the base is 101.5%, b=1.015

substitute

f(t)=300(1.015)^t

Part 6) A research laboratory is testing a new vaccine on 300 infected cells. The decay rate is 1.5% per minute. Find the number of infected cells after t minutes.

we know that

This question is about exponent function of the form

f(x)= a(b)^t

where

a is the initial value

b is the base of the exponent.

we have

A research laboratory is testing new vaccine on 300 infected cells

initial value is a=300

The decay/decrease rate is 1.5% per minute

Since it will decrease by 1.5% every min, it will become: 100%-1.5%=98.5%

the base is 98.5%, b=0.985