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podryga [215]
3 years ago
6

Ellie is a 45-year-old woman who has not yet planned for retirement. She would like to retire at age 67. Answer each question us

ing complete sentences.
Part I: In what type of investments should Ellie deposit her money? Why?
Part II: At what age, or point in your life, would you like to start saving for retirement?
Part III: In what type of investments would you like to deposit your money into when you start saving for retirement? Why?
Mathematics
1 answer:
timofeeve [1]3 years ago
6 0
I. Exchange traded funds (ETFs) equally distributed among the US stock fund, a US bond fund, an international stock fund, and an international bond fund. This is because ETFs provides diversification and low expenses.

ii.The  perfect time to start saving is as soon as you have a full-time job. If the employer has a tax deferred investment plan, and if the employer matches some of your investment then you'll be ready to get extra benefits.

iii. I'd put my investments in ETFs equally distributed among a US stock fund, an international fund, a US bond fund and an international bond fund. This reduces expenses through diversification. 
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Step-by-step explanation:

Put the numbers in place of the corresponding variables and do the arithmetic.

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Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

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