Question

1. suppose fernley deposited $5,000 in an interest-bearing account for his savings fund that earns simple interest at 3.5% each year. jasleen deposited $2,500 in an interest-bearing account for her savings fund that earns compound interest at 3.75% each year (+6 pts).

a) write the function equation for each relationship.
jasleens account:
fernleys account:
b) graph each relationship
c) determine the average rate of change between t = 0 and t = 6.

1. the number of people in the town of hooville was 32,645. a year later, the number of people was 43,937. write a function to model this situation and determine the percent increase or decrease of the population each year (+3 pts).

Explanation:

Problem 9

Part a)

Step1: Define Jasleen's compound interest function

The compound interest formula is $A = P(1 + r)^t$, where $P=2500$, $r=0.0375$.
$J(t) = 2500(1 + 0.0375)^t = 2500(1.0375)^t$

Step2: Define Fernley's simple interest function

The simple interest formula is $A = P(1 + rt)$, where $P=5000$, $r=0.035$.
$F(t) = 5000(1 + 0.035t) = 5000 + 175t$

Part c)

Jasleen's Account

Step1: Find $J(0)$ and $J(6)$

$J(0) = 2500(1.0375)^0 = 2500$
$J(6) = 2500(1.0375)^6 \approx 2500 \times 1.244 = 3110$

Step2: Calculate average rate of change

Average rate of change = $\frac{J(6)-J(0)}{6-0}$
$\frac{3110 - 2500}{6} = \frac{610}{6} \approx 101.67$

Fernley's Account

Step1: Find $F(0)$ and $F(6)$

$F(0) = 5000 + 175(0) = 5000$
$F(6) = 5000 + 175(6) = 5000 + 1050 = 6050$

Step2: Calculate average rate of change

Average rate of change = $\frac{F(6)-F(0)}{6-0}$
$\frac{6050 - 5000}{6} = \frac{1050}{6} = 175$

Step1: Define exponential growth function

The population growth model is $P(t) = P_0(1 + r)^t$, where $P_0=32645$.
First find $r$: $\frac{43937}{32645} = 1 + r$

Step2: Calculate growth rate $r$

$r = \frac{43937}{32645} - 1 \approx 1.346 - 1 = 0.346$
Convert to percentage: $0.346 \times 100 = 34.6\%$

Step3: Write the function

$P(t) = 32645(1.346)^t$

Answer:

Part a)

Jasleen's Account: $J(t) = 2500(1.0375)^t$
Fernley's Account: $F(t) = 5000 + 175t$

Part c)

Jasleen's Account average rate of change: $\approx \$101.67$ per year
Fernley's Account average rate of change: $\$175$ per year

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Problem 10