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sweet-ann [11.9K]
3 years ago
8

You have won the lottery and will receive 20 annual payments of $10,000 starting today. If you can invest these payments at 8.5%

, what is the present value of your winnings?
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

$102,677.20

Step-by-step explanation:

The present value of an annuity due is determined by the following expression:

PV = P+P*(\frac{1-(1+r)^{-n+1}}{r})

Where 'P' is the amount of each payment received, 'r' is the interest rate on the investment and 'n' is the number of yearly payments.

With 20 annual payments of $10,000 at a rate of 8.5%, the present value is:

PV = 10,000+10,000*(\frac{1-(1+0.085)^{-20+1}}{0.085})\\PV = 10,000+ 10,000*(9.26772)\\PV=\$102,677.20

The present value of your winnings is $102,677.20.

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An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
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Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

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∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

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