Mathematics, 22.02.2020 05:20 SmokeyRN

John, sally, 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.

1. What type of equation models Johns situation?

2. Write the model equation for Johns situation

3. How much money will John have after 2 years?

4. How much money will John have after 10 years?

5. What type of exponential model is Sally’s situation?

6. Write the model equation for Sally’s situation

7. How much money will sally have after 2 years?

8. How much money will sally have after 10 years?

9. What type of exponential model is Natalie’s situation?

10. Write the model equation for Natalie’s situation

11. How much money will Natalie have after 2 years?

12. How much money will Natalie have after 10 years?

13. Who will have the most money after 10 years?

Answers: 3

Show answers

Answers

Answer from: myapawesomeox22sq

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) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 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

$b=\$300$

y=100x+300

therefore

John’s situation is modeled by a linear equation

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

see part 1)

y=100x+300

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

Remember that

1 year is equal to 12 months

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

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

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) 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 6) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$

see the Part 5)

Part 7) 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 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) 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 10) Write the model equation for Natalie’s situation

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

see Part 9)

Part 11) 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 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) 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

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

Answer from: Quest

Idk idk idk idk idk idk idk idk

Answer from: Quest

Maybe fx = lxl because of the way the x is positioned is that what he

Another question on Mathematics

Mathematics, 21.06.2019 18:20

F(n + 1) = f(n) – 8. if f(1) = 100, what is f(6)?

Answers: 1

Answer

Mathematics, 21.06.2019 18:50

Find the greatest common factor of the followig monomials 21u^5v,3u^4v^6, and 3u^2v^5

Answers: 1

Answer

Mathematics, 21.06.2019 20:30

Secant be and cf intersect at point d inside a what is the measure of cde

Answers: 1

Answer

Mathematics, 22.06.2019 01:50

:i need some real : a store sells shirts to the public at one pricing scale and wholesale at another pricing scale. the tables below describe the cost, y, of x shirts. (after tables) how do the slopes of the lines created by each table compare? the slope of the public table is 3/4 of the slope of the wholesale table.the slope of the wholesale table is 3/4 of the slope of the public table.the slope of the public table is 12 times greater than the slope of the wholesale table.the slope of the wholesale table is 12 times greater than the slope of the public table.

Answers: 3

Answer

You know the right answer?

John, sally, Natalie would all like to save some money. John decides that it would be best to save m...

Questions