Answer :
Answer:
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) [tex]f(t)=7(2)^t[/tex]
Part 4) [tex]f(t)=3,000(0.93)^t[/tex]
Part 5) [tex]f(t)=300(1.015)^t[/tex]
Part 6) [tex]f(t)=300(0.985)^t[/tex]
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
[tex]I=P(rt)[/tex]
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
[tex]t=3\ years\\ P=\$1,000\\r=0.10[/tex]
substitute in the formula above
[tex]I=\$1,000(0.10*3)=\$300[/tex]
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
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
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
[tex]t=2\ years\\ P=\$1,000\\ r=0.05\\n=1[/tex]
substitute in the formula above
[tex]A=\$1,000(1+\frac{0.05}{1})^{1*2}[/tex]
[tex]A=\$1,000(1.05)^{2}=\$1,102.50[/tex]
The interest is equal to
[tex]I=\$1,102.50-\$1,000=\$102.50[/tex]
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
[tex]f(x)= a(b)^t[/tex]
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
[tex]f(t)= 7(2)^t[/tex]
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
[tex]f(x)= a(b)^t[/tex]
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
[tex]f(t)=3,000(0.93)^t[/tex]
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
[tex]f(x)= a(b)^t[/tex]
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
[tex]f(t)=300(1.015)^t[/tex]
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
[tex]f(x)= a(b)^t[/tex]
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
substitute
[tex]f(t)=300(0.985)^t[/tex]