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zhenek [66]
2 years ago
5

% interest compounded annually until Bob retires on his 65th birthday. How much is the IRA worth when Bob retires

Business
1 answer:
ira [324]2 years ago
6 0

Answer:

The worth of the IRA when Bob retires at 65 is $190,706.57.

Explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Bob makes his first $1,200 deposit into an IRA earning 6.5% compounded annually on the day he turns 24 and his last $1,200 deposit on the day he turns 44 (21 equal deposits in all.) With no additional deposits, the money in the IRA continues to earn 6.5% interest compounded annually until Bob retires on his 65th birthday. How much is the IRA worth when Bob retires?

The explanation of the answer is now given as follows:

Step 1: Calculation of the future value of the IRA when Bob turns 44

This can be calculated using the formula for calculating the Future Value (FV) of an Ordinary Annuity as follows:

FV44 = M * (((1 + r)^n - 1) / r) ................................. (1)

Where,

FV44 = Future value of the IRA when Bob turns 44 = ?

M = Annuity payment = $1,200

r = annual interest rate = 6.5%, or 0.065

n = number of years = 44 - 24 + 1 = 21

Substituting the values into equation (1), we have:

FV44 = $1,200 * (((1 + 0.065)^21 - 1) / 0.065)

FV44 = $1,200 * 42.3489537330236

FV44 = $50,818.74

Step 1: Calculation of the future value of IRA when Bob retires at 65

This can be calculated using the simple future value formula as follows:

FV65 = FV44 * (1 + r)^n ....................................... (1)

Where;

FV65 = Future value of IRA when Bob retires at 65 or the worth of the IRA when Bob retires at 65 = ?

FV44 = Future value of the IRA when Bob turns 44 = $50,818.74

r = annual interest rate = 6.5%, or 0.065

n = number of years = 65 - 44 = 21

Substituting the values into equation (2), we have:

FV65 = $50,818.74 * (1 + 0.065)^21

FV65 = $50,818.74 * 3.75268199264653

FV65 = $190,706.57

Therefore, the worth of the IRA when Bob retires at 65 is $190,706.57.

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Answer:

The percentage decrease in utilization is 83.33%

Explanation:

According to the data, we have the following:

Coefficient of variance, m = 3

Arrival rate, ra = 45 per hour

Service rate, re = 18 per hour per lane

Therefore, in order to calculate the percentage decrease in utilization when one more checkout lane is added to the system, we have to use the following formula:

So, percentage decrease in utilization = ra / (m.re)

                                                                = 45 / (3*18) = 0.833

The percentage decrease in utilization is 83.33%

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3 years ago
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Send an e-mail to all employees

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2 years ago
A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If
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Answer:

E = 1.20 × 10^{4} N/C

V = 1800 V

x = 0.058 m

Explanation:

given data

radius = 15 cm

net charge = 3 × 10^{-8} C

electric potential decreased = 500 V

solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =  \frac{q}{4\pi \epsilon _o R^2}   ............1

put here value

electric field at the sphere’s surface E  = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}    

E = 1.20 × 10^{4} N/C

and

potential on surface of sphere is

V =  \frac{q}{4\pi \epsilon _o R} ................2

V = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15}  

V = 1800 V

and

now we get distance that is x

and we know here

ΔV = V(x) - V   ..............3

substitute here value

-500V = \frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})

-500 V = {3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})

solve it we get x

x = 0.058 m

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2 years ago
Your sister is starting 9th grade next year and is thinking about going to college. What step would you recommend she take first
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Just before Henderson Laboratories opened for business, Eugene Henderson, the owner, had the following assets and liabilities. C
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Answer:

Assets = Laboratory Equipment ( Fixed asset) + Laboratory supplies (Current Asset) + Cash ( Current asset)

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= $275,600

Liabilities = Loan Payable ( Long term liability) + Accounts Payable ( current liability)

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Assets = Liabilities + Owners Equity

Owners Equity = Assets - Liabilities

= 275,600 - 53,150

= $222,450

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