PVHS Algebra 2 Final

1) Population Growth 1 In the population of 1931 Geekville was 11931. If the growth rate is 5.8%, what was the population in ? 1993

10494

434910

202718

139437

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Explanation

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2) Compound Interest 1 How much will you have in your account if: $57387 is invested at 11.2% compounded semi-annually for 84 years

$211637693.12

$542440089.7

$2184.39

$167350.18

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Explanation

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3) Radioactive Decay 2 You discovered a new radioactive isotope isotope and named it Geekonium-25. At 9 a.m., you have 74 mg in your petrie dish and at 7 p.m., you measure only 56 mg. What's the half-life?

Half-life = 60.62 hours

Half-life = 24.87 hours

Half-life = 20.65 hours

Half-life = 24.2 hours

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Explanation

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4) Radioactive Decay 2 You discovered a new radioactive isotope isotope and named it Geekonium-25. At 10 a.m., you have 31 mg in your petrie dish and at 2 p.m., you measure only 18 mg. What's the half-life?

Half-life = 8.54 hours

Half-life = 9.27 hours

Half-life = 4.85 hours

Half-life = 5.1 hours

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Explanation

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5) Compound Interest 1 How much will you have in your account if: $4865 is invested at 10.4% compounded monthly for 62 years

$68986.34

$65096.82

$2987779.59

$90372.21

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Explanation

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6) Radioactive Decay 1 The half-life of boogonium is 326 years. If you have 36 grams in 1923, how much will be left in 2080?

25.45 grams

34.64 grams

25.78 grams

18.21 grams

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Explanation

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7) Population Growth 1 In the population of 1925 Geekville was 1329. If the growth rate is 8.6%, what was the population in ? 1971

69437

44812812

132696

77889

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Explanation

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8) Radioactive Decay 1 The half-life of boogonium is 10 years. If you have 15 grams in 1930, how much will be left in 2053?

8.05 grams

0 grams

6.91 grams

18.21 grams

Since the half-life of boogonium is 10 years, it means that half of the substance will decay every 10 years. From 1930 to 2053, there is a time difference of 123 years, which is equivalent to 12.3 half-lives. Therefore, after 12.3 half-lives, there will be no boogonium left, resulting in 0 grams remaining.

Explanation

Since the half-life of boogonium is 10 years, it means that half of the substance will decay every 10 years. From 1930 to 2053, there is a time difference of 123 years, which is equivalent to 12.3 half-lives. Therefore, after 12.3 half-lives, there will be no boogonium left, resulting in 0 grams remaining.

9) Continuous Compound Interest 2 How long will it take: $49835 to grow into $85721 if it is invested at 10.6% compounded continuously

38.02 years

11.56 years

8.58 years

5.12 years

The correct answer is 5.12 years. This can be calculated using the formula for continuous compound interest: A = P * e^(rt), where A is the final amount, P is the initial principal, r is the interest rate, and t is the time in years. Rearranging the formula to solve for t, we have t = ln(A/P) / r. Plugging in the values given in the question, we get t = ln(85721/49835) / 0.106. Evaluating this expression gives t ≈ 5.12 years.

Explanation

The correct answer is 5.12 years. This can be calculated using the formula for continuous compound interest: A = P * e^(rt), where A is the final amount, P is the initial principal, r is the interest rate, and t is the time in years. Rearranging the formula to solve for t, we have t = ln(A/P) / r. Plugging in the values given in the question, we get t = ln(85721/49835) / 0.106. Evaluating this expression gives t ≈ 5.12 years.

10) Compound Interest 2 How long will it take: $27065 to grow into $88726 if it is invested at 2.3% compounded semi-annually

18.12 years

26.47 years

51.92 years

15.81 years

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Explanation

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11) Population Growth 1 In the population of 1931 Geekville was 29214. If the growth rate is 2.9%, what was the population in ? 1944

269870

577048

27657

42591

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Explanation

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12) Compound Interest 2 How long will it take: $42192 to grow into $80437 if it is invested at 6.4% compounded daily

15.8 years

9.23 years

7.48 years

10.08 years

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Explanation

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13) Continuous Compound Interest 2 How long will it take: $6574 to grow into $53287 if it is invested at 2.8% compounded continuously

25.23 years

12.39 years

74.73 years

50.79 years

The correct answer is 74.73 years. This can be calculated using the formula for continuous compound interest, which is A = P * e^(rt), where A is the final amount, P is the initial principal, e is Euler's number (approximately 2.71828), r is the interest rate, and t is the time in years. In this case, we have P = $6574, A = $53287, r = 2.8%, and we need to solve for t. Rearranging the formula, we get t = ln(A/P) / r. Plugging in the values, we get t = ln($53287/$6574) / 0.028, which is approximately equal to 74.73 years.

Explanation

The correct answer is 74.73 years. This can be calculated using the formula for continuous compound interest, which is A = P * e^(rt), where A is the final amount, P is the initial principal, e is Euler's number (approximately 2.71828), r is the interest rate, and t is the time in years. In this case, we have P = $6574, A = $53287, r = 2.8%, and we need to solve for t. Rearranging the formula, we get t = ln(A/P) / r. Plugging in the values, we get t = ln($53287/$6574) / 0.028, which is approximately equal to 74.73 years.

14) Radioactive Decay 2 You discovered a new radioactive isotope isotope and named it Geekonium-25. At 5 a.m., you have 85 mg in your petrie dish and at 6 p.m., you measure only 25 mg. What's the half-life?

Half-life = 7.36 hours

Half-life = 4.85 hours

Half-life = 4.35 hours

Half-life = 7.48 hours

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Explanation

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Pvhs Algebra 2 Final