For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

Using the formula $A(t)=P(1+rt)$, we can calculate A for any time $t$.

a.)

$P=500$ and $r=0.064$

$A(1)=500[1+(0.064)(1)]=$532$

$A(2)=500[1+(0.064)(2)]=$564$

$A(3)=500[1+(0.064)(3)]=$596$

$$
A(15)=500[1+(0.064)(15)]=$980
$$

b.)

$P=1250$ and $r=0.041$

$A(1)=1250[1+(0.041)(1)]=$1301.25$

$A(2)=1250[1+(0.041)(2)]=$1352.50$

$A(3)=1250[1+(0.041)(3)]=$1403.75$

$$
A(15)=500[1+(0.041)(15)]=$2018.75
$$

c.)

$P=25000$ and $r=0.05$

$A(1)=25000[1+(0.05)(1)]=$26;250$

$A(2)=25000[1+(0.05)(2)]=$27;500$

$A(3)=25000[1+(0.05)(3)]=$28;750$

$$
A(15)=500[1+(0.05)(15)]=$43;750
$$

d.)

$P=1700$ and $r=0.023$

$A(1)=25000[1+(0.023)(1)]=$1739.10$

$A(2)=25000[1+(0.023)(2)]=$1778.20$

$A(3)=25000[1+(0.023)(3)]=$1817.30$

$$
A(15)=500[1+(0.023)(15)]=$2286.50
$$

a.) $$980$

b.) $$2018.75$

c.) $43;750$

d.) $2286.50$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

a.) From the formula of $A(t)$, observe that when $t=0$, $A(t)=P$, thus, the intercept to the vertical axis should correspond to the principal amount $P$ which can be read from the graph as $$2000$.

b.) The interest after four years is $I=2500-2000=500$. Using the formula $I=Prt$, we can solve for $r$ as

$500=2000cdot rcdot 4$

$r=dfrac{500}{2000cdot 4}=0.0625=6.25%$

The interest after 5 years is therefore

$$
I=Prt=(2000)(0.0625)(5)=$625
$$

c.) As calculated in part(b), $r=6.25%$

d.) We shall the accumulated amount $A(t)$ as a function of $t$.

$A(t)=P(1+rt)$

$A(t)=2000(1+0.0625t)$

$$
A(t)=2000+125t
$$

a.) $P=$2000$

b.) $$625$

c.) $r=6.25%$

d.) $A(t)=2000+125t$

For simple interest, the formula is

$$
I = Prt
$$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

In this case, the given values are

$P = 850$

$I = 200$

$r = 0.07$

We must find the time so that $I=200$

$$
begin{gather*}
I = Prt\[0.5 em]
200 = (850)(0.07)(t)\[0.5 em]
t=dfrac{200}{(850)(0.07)}=bold{3.36;years}
end{gather*}
$$

$3.36$ years

In this case, the given values are

$P = 2845$

$t =dfrac{12}{365};$years

$I = 26.19$

We must find the annual interest rate $r$

$I = Prt$

$26.19=(2845)(r)left(dfrac{12}{365}right)$

$$
r=dfrac{26.19}{2845cdot dfrac{12}{365}}=boxed{bold{0.28; or; 28%}}
$$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

$$
r=28%
$$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

[begin{gathered}
{text{a}}{text{.)}} hfill \
P = 500 hfill \
r = 0.048 hfill \
t = 8 hfill \
hfill \
I = P cdot r cdot t = left( {500} right)left( {0.048} right)left( 8 right) = $ 192 hfill \
A = P + I = 500 + 192 = $ 692 hfill \
end{gathered} ]

[begin{gathered}
{text{b}}{text{.)}} hfill \
P = 3200 hfill \
r = 0.098 hfill \
t = 12 hfill \
I = P cdot r cdot t = left( {3200} right)left( {0.098} right)left( {12} right) = $ 3763.20 hfill \
A = P + I = 3200 + 3763.20 = $ 6963.20 hfill \
end{gathered} ]

[begin{gathered}
{text{c}}{text{.) }} hfill \
P = 5000 hfill \
r = 0.039 hfill \
t = frac{{16}}{{12}} hfill \
I = P cdot r cdot t = left( {5000} right)left( {0.039} right)left( {16/12} right) = $ 260 hfill \
A = 5000 + 260 = $ 5260 hfill \
end{gathered} ]

[begin{gathered}
{text{d}}{text{.)}} hfill \
P = 128 hfill \
r = 0.18 hfill \
t = 5/12 hfill \
I = P cdot r cdot t = left( {128} right)left( {0.18} right)left( {5/12} right) = $ 9.60 hfill \
A = P + I = 128 + 9.6 = $ 137.60 hfill \
end{gathered} ]

[begin{gathered}
{text{e}}{text{.)}}, hfill \
P = 50000 hfill \
r = 0.24 hfill \
t = 17/52 hfill \
I = P cdot r cdot t = left( {50000} right)left( {0.24} right)left( {17/52} right) = $ 3923.08 hfill \
A = P + I = 50000 + 3923.08 = $53923.08 hfill \
end{gathered} ]

[begin{gathered}
{text{f}}{text{.)}}, hfill \
P = 4500 hfill \
r = 0.12 hfill \
t = 100/365 hfill \
I = P cdot r cdot t = left( {4500} right)left( {0.12} right)left( {100/365} right) = $ 147.95 hfill \
A = P + I = 4500 + 147.95 = $ 4647.95 hfill \
end{gathered} ]

a.) $I=$192$ , $A=$692$

b.) $I=$3763$, $A=$6963.20$

c.) $I=$260$, $A=$5260$

d.) $I=$9.60$, $A=$137.60$

e.) $I=$3923.08$ , $A=$53923.08$

f.) $I=$147.95$ , $A=$4647.95$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

In this case, we need to find the annual interest rate given that an accumulated amount of $A(t)=$8000$ is obtained from an an initial amount of $P=4,800$ after $t=8.5 years$.

Using the formula

$A(t)=P(1+rt)$

$8000=4800(1+rcdot 8.5)$

$$
r=dfrac{(8000/4800)-1}{8.5}=0.0784=boxed{bold{7.84%}}
$$

7.84$%$.

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

Here we must find $P$ that will yield an interest of $I=$250$ per month $left(dfrac{1}{12}text{ of a year} right)$ at $r=6.3%=0.063$ annual interest.

We can use the formula $I=Prt$

$250=P(0.063)cdot dfrac{1}{12}$

$P=dfrac{250}{0.063cdot 1/12}=47;619.05$

Therefore, not less than $boxed{bold{$47;619.05}}$ must be invested to have an interest of $$250$ per month.

$$
$47;619.05
$$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

a.) We are given with initial investment of $P=3500$, an interest rate of $r=5.5%$. We shall find the amount of interest per 1 year using the formula $I=Prt$.

$I=(3,500)(0.055)(1)=192.5$

The account value increases by $$192.50$ every year.

b.) The amount for the first 5 years can be calculated using $A(t)=P(1+rt)$.

$A(1)=3500[1+0.055(1)]=$3692.50$

$A(2)=3500[1+0.055(2)]=$3885.00$

$A(3)=3500[1+0.055(3)]=$4077.50$

$A(4)=3500[1+0.055(4)]=$4270.00$

$A(5)=3500[1+0.055(5)]=$4462.50$

c.) The accumulated amount after $n$ years is therefore

$A(n)=3500(1+0.055n)=3500+192.5n$

a.) $$192.50$ each year

b.) 3692.50, 3885.00, 4077.50, 4270.00, 4462.50

c.) $A(n)=3500+192.5n$

d.) see graph

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

a.) For simple interests, the account will increase in value by a constant amount every quarter (0.25 year).

Therefore,

$A_2-A_1=A_3-A_2$

We know that $A_2=3994.32$, $A_3=4248.64$

$3994.32-A_1=4248.64-3994.32implies A_1=3740$

Therefore, the initial amount invested is $boxed{bold{$3740}}$.

We shall calculate the interest rate using

$I=Prt$

$4248.64-3994.32=(3740)(r)(0.25;text{years})$

$r=0.272=27.2%$ annually or $dfrac{27.2%}{4}=6.8%$ quarterly.

Therefore, the interest rate is $boxed{bold{27.2%;annually}}$.

a.) $3740$

b.) $27.2%$ annually

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

a.) For simple interests, the investment or debt will increase in value by a constant amount every year.

Therefore,

$A_2-A_1=A_3-A_2$

We know that $A_2=2081.25$, $A_3=2312.5$

$2081.25-A_1=2312.5-2081.25implies A_1=1850$

Therefore, the initial amount borrowed was $$1,850$.

b.) The interest that Anita needs to pay is also constant.

$$
I=A_3-A_2=2312.5-2081.25=$231.25
$$

c.) We shall calculate when will Anita have a debt of $$7500$. The debt increases by $231.25$ each year from an initial debt of $1850$. Therefore, the time can be calculated as

$7500=1850+231.25n$

$n=dfrac{7500-1850}{231.25}=24.4324$ years.

Since $(0.4324cdot 365)=157.8$

Anita’s debt would be $$7500$ after $boxed{bold{ 24; years ;and ;158 ;days}}$.

a.) $P=$1,850$

b.) $I=$231.25$

c.) 24 years and 158 days

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

We shall find the time when Len’s and Dave’s investment would be equal.

We will denote $A_L$ and $A_D$ as the account value of Len and Dave respectively.

Using the formula for the accumulated amount after $t$ years,

$A_L=5200(1+0.03t)$

$A_D=3600(1+0.05t)$

Equate the two equations

$5200(1+0.03t)=3600(1+0.05t)$

$5200+0.03(5200)t=3600+3600(0.05)t$

$[0.03(5200)-3600(0.05)]t=3600-5200$

$-24t=-1600$

$t=dfrac{200}{3}=66dfrac{2}{3}$ years

Therefore, Len’s and Dave’s account value would be equal after $boxed{bold{66; years; and; 8 ;months}}$.

Therefore the principal $P$ is 750

The general formula for simple interest is:

$$
A(t)=P+Prt
$$

$$
r=dfrac{27.75}{750}=0.037
$$

To calculate the rate divide the value being multiplied by $t$ by the principal

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

We shall find a formula for doubling time $t_d$, which is the time when the initial amount $P$ becomes $2P$.

The accumulated amount then should be $A(t)=2P$.

$2P=P(1+rcdot t_d)$

$2=1+rcdot t_dimplies t_d=dfrac{2-1}{r}$

$$
t_d=dfrac{1}{r}
$$

$$
dfrac{1}{r}
$$

For simple interest, the formula is

$I = Prt$

where

$I$ = simple interest

$P$ = principal amount or initial investment

$r$ = annual interest rate

$t$ = number of years

The accumulated amount after $t$ years A(t) is

$$
A(t)=P+I=P(1+rt)
$$

In this case, the principal amount changes every year as new investment is coming in.
At $r=6.4%=0.064$, the amount after 25 years due to the investment in the $n^{th}$ year which we will denote as $I_n$ is

$I_1=500(1+0.064cdot 25)=1300$

$I_2=500(1+0.064cdot 24)=1268$

$I_3=500(1+0.064cdot 23)=1236$

The total amount on her $25^{th}$ birthday is the sum of accumulated amount from the investments each year.

Thus, this is an arithmetic series with $t_1=1300$ and $d=1268-1300=-32$.

Since the first investment was made when Sara was born $(n=0)$, the total account value at her 25$^{th}$ birthday is the result of $26$ investments

$S_n=dfrac{n}{2}[2t_1+(n-1)(d)]$

$S_{26}=dfrac{26}{2}[2(1300)+(26-1)(-32)]=23400$

Therefore, on Sara’s 25th birthday, the account value would be $boxed{bold{$23;400}}$.

$$
$23,400
$$