Question

John, Sally, and Natalie would all like to save some money. John decides that it would be best to save money in a jar in his closet every single month. He decides to start with $300, and then save $100 each month. Sally has $6000 and decides to put her money in the bank in an account that has a 7% interest rate that is compounded annually. Natalie has $5000 and decides to put her money in the bank in an account that has a 10% interest rate that is compounded continuously
(1pt) What type of equation models John’s situation?_________________ (1pt) Write the model equation for John’s situation___________________
(1pt) How much money will John have after 10 years? ________________ (1pt) What type of exponential model is Sally’s situation? ______________
(1pt) Write the model equation for Sally’s situation ___________________
(1pt) How much money will Sally have after 10 years? ________________ (1pt) What type of exponential model is Natalie’s situation? ____________
(1pt) Write the model equation for Natalie’s situation _________________
(1pt) How much money will Natalie have after 10 years? ______________ (1pt) Who will have the most money after 10 years? ________________

Answer 1

Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\ 12,300$

Part 4) Is a exponential growth function

Part 5) $A=6,000(1.07)^{t}$  

Part 6) $\ 11,802.91$  

Part 7) Is a exponential growth function

Part 8) $A=5,000(e)^{0.10t}$    or  $A=5,000(1.1052)^{t}$  

Part 9)  $\ 13,591.41$

Part 10) Natalie has the most money after 10 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

$y=mx+b$

The slope is equal to

$m=\ 100\ per\ month$

The y-intercept or initial value is

$b=\ 300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

$y=100x+300$

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

10 years=10(12)=120 months

For x=120 months

substitute in the linear equation

$y=100(120)+300=\ 12,300$

Part 4) What type of exponential model is Sally’s situation?

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  

$P=\ 6,000\\ r=7\%=0.07\\n=1$  

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}$  

$A=6,000(1.07)^{t}$  

therefore

Is a exponential growth function

Part 5) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$  

see the Part 4)

Part 6) How much money will Sally have after 10 years?

For t=10 years

substitute  the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\ 11,802.91$  

Part 7) What type of exponential model is Natalie’s situation?

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  

$P=\ 5,000\\r=10\%=0.10$  

substitute in the formula above

$A=5,000(e)^{0.10t}$  

Applying property of exponents

$A=5,000(1.1052)^{t}$  

therefore

Is a exponential growth function

Part 8) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$    or  $A=5,000(1.1052)^{t}$

see Part 7)

Part 9) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\ 13,591.41$

Part 10) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Answer 2

Answer:

See answer below

Step-by-step explanation:

What type of equation models John’s situation? linar/exponential (both linear and exponential)

Write the model equation for John’s situation  y = 300+100(.001)^(number of months)

How much money will John have after 2 years? $789.89

How much money will John have after 10 years? $1395.44

What type of exponential model is Sally’s situation? exponential upwards

Write the model equation for Sally’s situation y = 6000(1.007)^(number of months/years)

How much money will Sally have after 2 years? $4000

How much money will Sally have after 10 years? $14,003.98

What type of exponential model is Natalie’s situation? complete interest

Write the model equation for Natalie’s situation: y = 5000(.10)^(number of years/months)

How much money will Natalie have after 2 years? $7,395.21

How much money will Natalie have after 10 years? $35,355.33

Who will have the most money after 10 years? This is a trick question. It looks like Natalie, but it's actually John. John is the one who worked hard to set aside the most money over the course of 10 years.