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Algebra II : Algebra II

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Example Questions

Example Question #1 : Interest Equations

Martisha invests $2000 into an account with continuously compounded interest. The account has an interest rate of 2.5%. Find the balance of the account after 2 years, rounded to the nearest cent.

Possible Answers:

$\$3297.44$

$\$296826.32$

$\$2102.54$

$\$2050.63$

Correct answer:

$\$2102.54$

Explanation:

To find the balance, B, of a continuously compounded interest account after a certain amount of time, we must use the following formula:

$B=Pe^{rt}$ , where P is the initial investment, r is the interest rate (as a decimal), and t is the amount of time being considered.

Plugging in all of the given information, we get

$B=2000e^{0.025\cdot 2}=2102.542193$

which rounded becomes $2102.54

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Example Question #9 : Interest Equations

How long will it take for Nikki to triple her initial investment into a continuously compounded interest account with an interest rate of 1.9%?

Possible Answers:

We are not given enough information to solve the problem

57.82 years

0.214 years

0.5782 years

Correct answer:

57.82 years

Explanation:

The formula to find the balance, B, of a continuously compounded interest account with interest rate, r, after a certain time, t, is given by

$B=Pe^{rt}$

To solve this problem, we need to know only the initial investment (P), our final balance (three times P) and the interest rate (expressed as a decimal), 0.019.

Plugging in our known information into the formula for continuously compounded interest, we get

$3P=Pe^{0.019t}$

We now solve for t:

$3=e^{0.019t}$

$\ln(3)=\ln(e^{0.019t})$

Exponentiating both sides allows us to get rid of the exponential:

$\ln3=0.019t$

t=57.82

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Example Question #311 : Exponents

Sheila wants to double her initial investment into a compounded interest account, with an interest rate of 4%. How long will this take, if the interest is compounded annually?

Possible Answers:

0.52 years

1.923 years

0.0565 years

17.67 years

Correct answer:

17.67 years

Explanation:

To determine the amount of time needed to double the initial investment - P - into a compound interest account, we simply plug in our given information into the formula:

$B=P(1+\frac{r}{n})^{nt}$

where B is the balance, P is the initial investment, r is the interest rate (as a decimal), n is the number of times the interest is compounded, and t is the time elapsed.

Now, because we are doubling P, our balance B becomes two times P:

$2P=P(1+\frac{0.04}{1})^{1\cdot t}$

Now, we can solve for P:

2=1.04^t

To bring the time variable down from being an exponent, we take the logarithm of both sides (common or natural):

$\ln2=t\cdot \ln1.04$

t=17.67

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Example Question #312 : Exponents

If a person deposits 300 dollars to a savings account, which earns one percent interest that is compounded annually, what is the balance after 60 years?

Possible Answers:

$\$5680$

$\$480$

$\$2540$

$\$545$

$\$385$

Correct answer:

$\$545$

Explanation:

Write the formula for compound interest.

$A=P(1+\frac{r}{n})^{nt}$

$A= \textup{balance}$

$P=\textup{starting investment or principal} = \$300$

$r=\textup{ rate} = 1\% = 0.01$

$n= \textup{number of times compounded per year} =1$

$t= \textup{time in years} = 60$

Substitute all the known values into the formula.

$A=300(1+\frac{0.01}{1})^{(1)(60)} = 300(1.01)^{60} = \$545$

The answer is: $\$545$

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Example Question #313 : Exponents

Suppose Billy's has $\$300$ , invests the money at a bank at $1\%$ , compounded monthly. About how much will Billy have after 36 months?

Possible Answers:

$\$323$

$\$318$

$\$309$

$\$303$

$\$333$

Correct answer:

$\$309$

Explanation:

Write the compound interest formula.

$A=P(1+\frac{r}{n})^{nt}$

where is the total, is the principal, is the rate, is the number of times compounded annually, and is the time in years.

$P=\$300$

$r=1\% = 0.01$

n=12

$t=\frac{36}{12}=3$

Substitute all into the equation.

$A=300(1+\frac{0.01}{12})^{12\times 3}$

$A=309.132 = \$309$

The answer is: $\$309$

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Example Question #314 : Exponents

Peter opens a savings account on his $21^{st}$ t birthday. He makes a deposit of $\$5,000$ . The account earns percent interest, compounded annually. Peter plans to take the money out when he is years old. If he doesn't make any deposits or withdrawals until then, how much money will be in the account?

Possible Answers:

$\$10,150.00$

$\$ 147,285.13$

$\$ 35,571.29$

$\$15,150.00$

Correct answer:

$\$ 35,571.29$

Explanation:

The formula for calculating compount interest is as follows:

$FV = PV\cdot (1+r)^n$

where

= future value

= present value

= interest rate

= number of times the interest is compounded

In this problem, the present value of the money is $5000, and the interest rate is 7%. If Peter takes the money out when he is 50, it would have been compounded 29 times (once per year). Therefore:

$FV = \$5000\cdot (1+.07)^{29}$

${\color{Red} FV = \$35,571.29}$

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Example Question #1 : Radioactive Decay Equations

Over the past few years, the number of students enrolled at a certain university has been decreasing. Each year there is a 12% decrease in student enrollement. Currently, 14,286 students are enrolled. If this trend continues, how many students will be enrolled in 5 years?

Possible Answers:

12,426

5,714

3,571

9,756

7,539

Correct answer:

7,539

Explanation:

This is an exponential decay problem. The formula for exponential decay is:

FV = PV(1- d)^n

Where

= future value

= present value

= rate of decay

= number of periods

This problem requests the number of students five years in the future. The rate of decay is twelve percent. Therefore:

FV = 14,286(1-.12)^5

FV = 7,539

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Example Question #2 : Radioactive Decay Equations

Sceintists recently discovered a new type of metal compound. They have roughly 15 grams of this compound, which has a half life of 16 hours. Approximately how much of this substance will the scientists have in 24 hours?

Possible Answers:

5.00 grams

0.06 grams

grams

0.50 grams

0.005 grams

Correct answer:

0.06 grams

Explanation:

Recall the radioactive decay formula:

$Q_{(t)}=Q_{(o)} (^{-r\cdot t})$

The half life formula is:

$r=\frac{ln2}{(h)}$ , where is the half life.

Plug in the given half life:

$r=\frac{ln2}{(16)}\approx 0.0433216$

Plug this value into the radioactive decay formula:

$Q_{(24)}=15 (^{-0.0433216\cdot 24})\approx 0.06$ grams

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Example Question #1 : Radioactive Decay Equations

The equation for radioactive decay is,

$A=A_{0}\cdot \left (\frac{1}{2} \right )^{t/h}$ .

Where $A_{0}$ is the original amount of a radioactive substance, is the final amount, is the half life of the substance, and is time.

The half life of Carbon-14 is about 5730 years. If a fossile contains grams of Carbon-14 at time t = 0 , how much Carbon-14 remains at time t = 17190 years?

Possible Answers:

$16 \text{ grams}$

$4 \text{ grams}$

$8 \text{ grams}$

None of the other answers.

$96 \text{ grams}$

Correct answer:

$8 \text{ grams}$

Explanation:

Using the equation for radioactive decay, we get:

$A=64\cdot \left (\frac{1}{2} \right )^{17190/5730}$ .

$A = 64\cdot \left ( \frac{1}{2} \right )^{3} = 64 \cdot \frac{1}{8} = 8 \text{ grams}$

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Example Question #3 : Radioactive Decay Equations

The number of fish in an aquarium is decreasing with exponential decay. The population of fish is decreasing by $7\%$ each year. There are 1500 fish in the aquarium today. If the decay continues how many fish will be in the aquarium in years?

Possible Answers:

990

None of these answers are correct

1122

1200

Correct answer:

1122

Explanation:

Every year the population of fish losses 7%. In other words, every year 93% of the fish remain from the previous year. Knowing this, we can use the original number of fish to find the number of fish for the next year. Since we want to know the number of fish 4 years from now, we multiply 1500 by 93% four times.

$\small \small 1500\cdot0.93\cdot0.93\cdot0.93\cdot0.93 = 1500\cdot(0.93)^4 \approx 1122$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Interest Equations

Example Question #9 : Interest Equations

Example Question #311 : Exponents

Example Question #312 : Exponents

Example Question #313 : Exponents

Example Question #314 : Exponents

Example Question #1 : Radioactive Decay Equations

Example Question #2 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

Example Question #3 : Radioactive Decay Equations

All Algebra II Resources

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