1. A car begins to depreciate at a rate of 24.9% annually as soon as it is driven off the lot. If a car was purchased for 26,500; how much is it worth after the second year? And what equation did you use? 2. Scientists studying a bacteria sample find that after starting with 60 cells, the number of cells doubles every hour. How many cells are there after 1 day? And what equation did you use? 3.You bought a car that was $25500 and the value depreciates by 4.5% each year. How much would the car be worth after 5 years?And what equation did you use? Also how much will the car be worth after eight years. And what is the equation you used to find how much it was worth for eight years.

1. Future value after two years = $14946.03

2. Number of bacteria after one day = 1006632960

3a. Future value after 5 years = $20256.16

3b. Future value after 8 years = $17642.79

Step-by-step explanation:

1. A car begins to depreciate at a rate of 24.9% annually as soon as it is driven off the lot. If a car was purchased for 26,500; how much is it worth after the second year? And what equation did you use?

Equation required = PR^n

P = initial value, 26500

R = depreciation ratio , (100-24.9)% = 0.751

n =  number of years elapsed, 2

Future Value of car after two years

FV = 26500*0.751^2 = 14946.03

2. Scientists studying a bacteria sample find that after starting with 60 cells, the number of cells doubles every hour. How many cells are there after 1 day? And what equation did you use?

Equation used: N(n) = n0 * (2)^n

N(n) = number of bacteria after n hours

n = 24 hours (1 day)

n0 = initial number of bacteria, 60

After 1 day (24 hours),

N(24) = 60* (2^24)

= 1006632960

(assuming sufficient food)

3.You bought a car that was $25500 and the value depreciates by 4.5% each year. How much would the car be worth after 5 years?And what equation did you use? Also how much will the car be worth after eight years. And what is the equation you used to find how much it was worth for eight years.

Using same equation as question 1,

P = 25500

3a.

n = 5

R = 1-0.045 = 0.955

Future value after 5 years

FV = 25500*0.955^5

= 20256.16

3b. n = 8

R = 1-0.045 = 0.955

Future value after 8 years

FV = 25500*0.955^8

= 17642.79

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step-by-step explanation: