Math  /  Algebra

QuestionFor each scenario below, find the matching growth or decay model, f(t)f(t). $100\$ 100 million dollars is invested in a compound interest account. The interest rate is 5\%, compounded every half a year. \qquad The concentration of pollutants in a lake is initially 100 ppm. The concentration decays by 30%30 \% every 3 years. The concentration of pollutants in a lake is initially 100 ppm. The concentration decays by 70\% every 3 years. 100 bacteria begin a colony in a petri dish. The bacteria increase by 30%30 \% every 3 hours. 100 bacteria begin a colony in a petri dish. The bacteria increase by 200\% every half hour. \square The cost of producing high end shoes is currently $100\$ 100. The cost is increasing by 50%50 \% every two years.
1. f(t)=100(0.7t3)f(t)=100\left(0.7^{\frac{t}{3}}\right)
2. f(t)=100(0.3t3)f(t)=100\left(0.3^{\frac{t}{3}}\right)
3. f(t)=100(1.3t3)f(t)=100\left(1.3^{\frac{t}{3}}\right)
4. f(t)=100(32t)f(t)=100\left(3^{2 t}\right)
5. f(t)=100(1.5t2)f(t)=100\left(1.5^{\frac{t}{2}}\right)
6. f(t)=100(1.052t)f(t)=100\left(1.05^{2 t}\right)

Studdy Solution

STEP 1

What is this asking? Match each money or bacteria situation with the right formula that tells us how much money or bacteria we have after some time. Watch out! Don't mix up growth and decay, and pay close attention to the time periods!

STEP 2

1. Compound Interest Account
2. Pollutant Decay (30%)
3. Pollutant Decay (70%)
4. Bacteria Growth (30%)
5. Bacteria Growth (200%)
6. Shoe Cost Increase

STEP 3

We're starting with $100\$100 million, and it grows by 5%5\% every half a year.
This means every half year we multiply by 1+0.05=1.051 + 0.05 = 1.05.

STEP 4

Since this happens twice a year, after tt years, we've compounded 2t2t times.
So, our formula is f(t)=100(1.052t)f(t) = 100(1.05^{2t}), which matches option 6.

STEP 5

We begin with 100100 ppm of pollutants.
A 30%30\% decay every 3 years means we multiply by 10.3=0.71 - 0.3 = 0.7 every 3 years.

STEP 6

If tt is in years, then the number of 3-year periods that have passed is t3\frac{t}{3}.
This gives us f(t)=100(0.7t3)f(t) = 100(0.7^{\frac{t}{3}}), matching option 1.

STEP 7

Starting at 100100 ppm, a 70%70\% decay every 3 years means we multiply by 10.7=0.31 - 0.7 = 0.3 every 3 years.

STEP 8

With tt in years, the number of 3-year periods is t3\frac{t}{3}.
So, our formula is f(t)=100(0.3t3)f(t) = 100(0.3^{\frac{t}{3}}), which is option 2.

STEP 9

We start with 100100 bacteria, growing by 30%30\% every 3 hours.
This means multiplying by 1+0.3=1.31 + 0.3 = 1.3 every 3 hours.

STEP 10

If tt is in hours, then we have t3\frac{t}{3} three-hour periods.
Our formula is f(t)=100(1.3t3)f(t) = 100(1.3^{\frac{t}{3}}), which is option 3.

STEP 11

Starting with 100100 bacteria, a 200%200\% increase every half hour means multiplying by 1+2=31 + 2 = 3 every half hour.

STEP 12

With tt in hours, we have 2t2t half-hour periods.
So, f(t)=100(32t)f(t) = 100(3^{2t}), which matches option 4.

STEP 13

The initial cost is $100\$100, increasing by 50%50\% every two years.
This means multiplying by 1+0.5=1.51 + 0.5 = 1.5 every 2 years.

STEP 14

If tt is in years, we have t2\frac{t}{2} two-year periods.
Thus, f(t)=100(1.5t2)f(t) = 100(1.5^{\frac{t}{2}}), matching option 5.

STEP 15

1. f(t)=100(1.052t)f(t)=100(1.05^{2t})
2. f(t)=100(0.7t3)f(t)=100(0.7^{\frac{t}{3}})
3. f(t)=100(0.3t3)f(t)=100(0.3^{\frac{t}{3}})
4. f(t)=100(1.3t3)f(t)=100(1.3^{\frac{t}{3}})
5. f(t)=100(32t)f(t)=100(3^{2t})
6. f(t)=100(1.5t2)f(t)=100(1.5^{\frac{t}{2}})

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