Please Help!
Jerry, Jack and Sophie are all hoping to save money! Jerry thinks saving money in a shoe box in his closet every month is a good idea. He decides to start with $125, and then save $50 each month. Jack was given $3520 from his
Grandma, and decides to put the money
into an account that has a 6.5% interest rate that is compounded annually. Sophie has earned $3500 working at the movie theater decides to put her money in the bank in an account that has a 7.05% interest rate that is compounded
continuously
Part 1: Describe the type of equation that models Jerry's situation. Create that equation of Jerry's situation. Using the equation you created, how much money will be in Jerry's account after 3 years? 10 years?
Part 2: Describe the type of equation that models Jack's situation. Create that equation of Jack's situation. Using the equation you created, how much mone
will be in Jack's account after 3 years? 10 years?
Part 3: Describe the type of equation that models Sophie's situation. Create that equation of Sophie's situation. Using the equation you created, how much money will be in Sophie's account after 3 years? 10 years?
Jerry's plan:
arithmetic series with a = 125 ,d = 50
part 1, n = 36
term(36) = a + 35d
= 125+35(50) = 1875
part 2, n = 120
.....
Jack:
Amount after 3 years = 3520(1 + .065)^3 = 4,251.98
amount after 10 years = .....
Sophie:
amount after 3 years = a e^(rt)
3520 e^(.07(3)) = 4342.55
amount after 10 years
= .....
Part 1:
Jerry's situation can be modeled by a linear equation, as he saves a fixed amount every month. The equation can be written as:
J(t) = 125 + 50t
Where J(t) represents the amount of money in Jerry's account after t months.
To calculate how much money will be in Jerry's account after 3 years (36 months), we substitute t = 36 into the equation:
J(36) = 125 + 50(36) = 125 + 1800 = $1925
To calculate how much money will be in Jerry's account after 10 years (120 months), we substitute t = 120 into the equation:
J(120) = 125 + 50(120) = 125 + 6000 = $6125
Part 2:
Jack's situation can be modeled by an exponential equation, as his money grows with an annual compounding interest rate. The equation can be written as:
J(t) = 3520(1 + 0.065)^t
Where J(t) represents the amount of money in Jack's account after t years.
To calculate how much money will be in Jack's account after 3 years, we substitute t = 3 into the equation:
J(3) = 3520(1 + 0.065)^3 = 3520(1.065)^3 ≈ $4016.17
To calculate how much money will be in Jack's account after 10 years, we substitute t = 10 into the equation:
J(10) = 3520(1 + 0.065)^10 = 3520(1.065)^10 ≈ $5964.06
Part 3:
Sophie's situation can be modeled by a continuous compound interest equation, as her money is compounded continuously. The equation can be written as:
S(t) = 3500e^(0.0705t)
Where S(t) represents the amount of money in Sophie's account after t years.
To calculate how much money will be in Sophie's account after 3 years, we substitute t = 3 into the equation:
S(3) = 3500e^(0.0705(3)) ≈ $4369.69
To calculate how much money will be in Sophie's account after 10 years, we substitute t = 10 into the equation:
S(10) = 3500e^(0.0705(10)) ≈ $5834.52
Now that we've covered the math, how about a joke to lighten the mood?
Why don't scientists trust atoms?
Because they make up everything!
Is there anything else to this problem I need the help
Part 1: Jerry's situation involves a simple linear equation, as he saves a fixed amount of money each month. The equation can be represented as:
Jerry's savings = initial amount + (monthly savings * number of months)
The initial amount Jerry starts with is $125, and he saves $50 each month. Assuming "t" represents the number of months, the equation can be written as:
Jerry's savings = $125 + ($50 * t)
To find how much money will be in Jerry's account after 3 years (36 months), we substitute t = 36 into the equation:
Jerry's savings = $125 + ($50 * 36)
Jerry's savings = $125 + $1800
Jerry's savings = $1925
Therefore, Jerry will have $1925 in his account after 3 years.
For 10 years (120 months):
Jerry's savings = $125 + ($50 * 120)
Jerry's savings = $125 + $6000
Jerry's savings = $6125
So, Jerry will have $6125 in his account after 10 years.
Part 2: Jack's situation involves compound interest, which can be represented by the formula:
Future Value = Principal * (1 + (interest rate / n))^(n * time)
Here, Principal refers to the initial amount of money, interest rate is the percentage rate (6.5% = 0.065), n is the number of times compounding occurs in a year (annually, so n = 1), and time is the number of years.
The equation for Jack's situation can be written as:
Jack's savings = Principal * (1 + (interest rate / n))^(n * time)
Jack's savings = $3520 * (1 + (0.065 / 1))^(1 * t)
For 3 years, we substitute t = 3 into the equation:
Jack's savings = $3520 * (1 + (0.065 / 1))^(1 * 3)
Jack's savings = $3520 * (1 + (0.065))^3
Jack's savings = $3520 * (1.065)^3
Jack's savings = $3956.87
Therefore, Jack will have $3956.87 in his account after 3 years.
For 10 years:
Jack's savings = $3520 * (1 + (0.065))^10
Jack's savings = $3520 * (1.065)^10
Jack's savings = $5377.28
So, Jack will have $5377.28 in his account after 10 years.
Part 3: Sophie's situation involves continuous compounding, which is represented by the formula:
Future Value = Principal * e^(interest rate * time)
Here, Principal refers to the initial amount of money, interest rate is the percentage rate (7.05% = 0.0705), e is the base of the natural logarithm (approximately 2.71828), and time is the number of years.
The equation for Sophie's situation can be written as:
Sophie's savings = Principal * e^(interest rate * time)
Sophie's savings = $3500 * e^(0.0705 * t)
For 3 years, we substitute t = 3 into the equation:
Sophie's savings = $3500 * e^(0.0705 * 3)
Sophie's savings = $3500 * e^(0.2115)
Sophie's savings = $4309.10
Therefore, Sophie will have $4309.10 in her account after 3 years.
For 10 years:
Sophie's savings = $3500 * e^(0.0705 * 10)
Sophie's savings = $3500 * e^(0.705)
Sophie's savings = $5726.62
So, Sophie will have $5726.62 in her account after 10 years.