Question

The balance y (in dollars) of your savings account after t years is represented by y=300(1.05)'. The beginning balance of your friend's account is $340, and the balance increases by $10 each year. a. Compare the account balances by calculating and interpreting the average rates of change from t=2 to t=6.​

Answer 1

To compare the account balances, calculate the average rates of change for both accounts from t=2 to t=6. The average rate of change for the first account is 15.97 dollars per year, and the average rate of change for the second account is 10 dollars per year.

Step-by-step explanation:

To compare the account balances, we need to calculate the average rates of change for both accounts from t=2 to t=6.

For the first account, the balance is given by the formula y = 300(1.05)^t. Substituting t=2 and t=6 into the formula, we get y = 300(1.05)^2 = 330.75 and y = 300(1.05)^6 = 391.61.

For the second account, the balance increases by $10 each year. So the balance at t=2 is $340 + 2*$10 = $360, and the balance at t=6 is $340 + 6*$10 = $400.

Therefore, the average rate of change for the first account is (391.61 - 330.75)/(6-2) = 15.97 dollars per year, and the average rate of change for the second account is (400 - 360)/(6-2) = 10 dollars per year.

Answer 2

My balance grows faster than my friends balance.

Step-by-step explanation:

My balance

`y=300(1.05)^t`

where

t = Time

Average rate from t=2 to t=6 is given by

`(t(6)-t(2))/(6-2)\\ =(300(1.05)^6-300(1.05)^2)/(4)\\ =\mathbf{17.82}`

My friend's balance is given by

`y=340+10t`

Average rate is given by

`(t(6)-t(2))/(6-2)\\ =((340+10* 6)-(340+10* 2))/(4)\\ =\mathbf{10}`

My balance grows faster than my friends balance.