3. The correct answer is:
d) f(x) = 175(1.22)^x; 473
To find the approximate population after 5 years, we can plug in x = 5:
f(5) = 175(1.22)^5 ≈ 472.91 ≈ 473
4. The correct answer is:
c) 7 years
To find out how many years it will take for the account balance to reach at least $6,000, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, t is the time in years.
Since the interest is compounded annually, n = 1. We can plug in the values and solve for t:
6,000 = 5,000(1 + 0.03/1)^(1*t)
1.2 = (1.03)^t
Now we'll need to use logarithms to solve for t:
t = ln(1.2) / ln(1.03) ≈ 6.12
Since we need the account balance to be at least $6,000, we round up the answer to 7 years.
3. An initial population of 175 quail increases at an annual rate of 22%. Write an exponential function to model the quail population. What will the approximate population be after 5 years?
a) f(x) = 175(0.22)^x; 473
b) f(x) = (175*0.22)^x; 84,587,005
c) f(x) = 175(22)^x; 901,885, 600
d) f(x) = 175(1.22)^x; 473
4. Suppose you deposit $5,000 in savings account that earns 3% annual interest. If you make no other withdrawals or depostis, how many years will it take the account balance to reach at least $6,000?
a) 10 years
b) 6 years
c) 7 years
d) 4 years
1 answer