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What will the value of an account (to the nearest cent) be after 8 years if \$100 is invested
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# Question : What will the value of an account (to the nearest cent) be after 8 years if \$100 is invested : 2158387

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Provide an appropriate response.

1) Find x to two decimal places.

x = 7,000e0.11

A) 7975.01

B) 7813.95

C) 7831.95

D) 8320.50

2) Find t to four decimal places.

e-t = 0.06

A) -2.8134

B) 2.9134

C) 2.8134

D) 2.6134

3) Find t to four decimal places.

e-0.07t = 0.05

A) -66.4815

B) 42.7962

C) -70.1312

D) 44.321

4) Graph the function which calculates the present value of an amount of \$5000 at an annual nominal rate of 7% compounded continuously for 0 ≤ t ≤ 10. Use the formula P = Ae-rt. A) B) C) D)  5) Find: 5000e-0.07t

A) 5000

B) 1

C) 0

D) ∞

6) A man with \$9000 to invest puts the money into an account that earns 8% compounded continuously. Graph the corresponding present value function and calculate the number of years before the \$9000 will be due in order for its present value to be \$7000. Use the formula P = Ae-rt. A) 3.14 years

B) 7.45 years

C) 0.84 years

D) 6.28 years

Solve the problem.

7) What will the value of an account (to the nearest cent) be after 8 years if \$100 is invested at 6.0% interest compounded continuously?

A) \$849.47

B) \$159.38

C) \$161.61

D) \$175.32

8) If \$5000 is invested at 5.25% compounded continuously, what is the amount in the account after 10 years?

A) \$7625.00

B) \$8452.29

C) \$7420.65

D) \$8442.52

9) How long will it take for the value of an account to be \$890 if \$350 is deposited at 11% interest compounded continuously? Round your answer to the nearest hundredth.

A) 8.48 yr

B) 9.33 yr

C) 10.41 yr

D) 0.93 yr

10) How long will it take for \$8400 to grow to \$14.600 at an interest rate of 9.4% if the interest is compounded continuously? Round the number of years to the nearest hundredth.

A) 5.88 yr

B) 0.06 yr

C) 58.81 yr

D) 0.59 yr

11) Suppose that \$8000 is invested at an interest rate of 5.5% per year, compounded continuously. How long would it take to double the investment?

A) 2 yr

B) 12.6 yr

C) 13.6 yr

D) 11.6 yr

12) How long will it take money to double if it is invested at 5.25%, compounded continuously? Round your answer to the nearest tenth.

A) 26.4 yr

B) 0.13 yr

C) 14 yr

D) 13.2 yr

13) An investor buys 100 shares of a stock for \$20,000. After 5 years the stock is sold for \$32,000. If interest is compounded continuously, what annual nominal rate of interest did the original \$20,000 investment earn? (Represent the answer as a percent to three decimal places.)

A) 1.200%

B) 9.400%

C) 8.470%

D) 0.094%

14) Radioactive carbon-14 has a continuous compound rate of decay of r = -0.000124. Estimate the age of a skull uncovered at an archaeological site if 6% of the original amount of carbon-14 is still present. (Compute answer to the nearest year.)

A) 124,027 yr

B) 20,032 yr

C) 470 yr

D) 22,689 yr

Find f'(x).

15) f(x) = 9ex - 4x + 2

A) 9ex - 2

B) 9xex-1 - 4

C) 9ex - 4

D) 9ex - 4x

16) f(x) = -6ex + 7x - 4

A) -6xex-1 + 7

B) -6ex + 7x

C) -6ex + 3

D) -6ex + 7

17) f(x) = x8 + 5ex

A) 8x7 + 5xex-1

B) 8x + 5ex

C) 8x7 + ex

D) 8x7 + 5ex

18) f(x) = 7ex - 3x2

A) 7xex-1 - 6x2

B) 7ex - 6x

C) 7ex - 2x

D) 7ex - 6x2

19) f(x) = -9lnx - x4 + 1

A) - (9/x) - 4x3

B) - (1/9x) - 4x3

C) - (9/x) - 4x

D) (9/x) - 4x3

20) f(x) = lnx4

A) 4lnx3

B) (4/x3)

C) (4/x)

D) (1/4x)

21) f(x) = lnx4 - 4x2

A) (4/x) - 4x

B) (1/4x) - 8x

C) (4/x) - 8x

D) (4/x3) - 8x

22) f(x) = lnx5 - 4ex + 2x2

A) (5/x) - 4xex-1 + 4x

B) (5/x4) - 4ex + 4

C) (5/x) - 4ex + 4x

D) (5/x) - 4ex + 2x

23) f(x) = 4lnx + lnx7 + 2ex

A) (4/x) + (7/x6) + 2xex-1

B) (11/x) + 2xex-1

C) (4/x) + (7/x6) + 2ex

D) (11/x) + 2ex

24) f(x) = 8ex + 4lnx3

A) 8ex + (12/x3)

B) 8ex + (12/x)

C) 8ex + (12/x2)

D) 8ex + (4/x2)

## Solution 5 (1 Ratings )

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Management Information Systems 1 Month Ago 144 Views