Question 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? (1 point) Responses ƒ(x) = 175(0.22)x; 473 ƒ( x ) = 175(0.22) x ; 473 ƒ(x) = (175 • 0.22)x; 84,587,005 ƒ( x ) = (175 • 0.22) x ; 84,587,005 ƒ(x) = 175(22)x; 901,885,600 ƒ( x ) = 175(22) x ; 901,885,600 ƒ(x) = 175(1.22)x; 473 ƒ( x ) = 175(1.22) x ; 473 Skip to navigation Item 1 , Fully Attempted. Item 2 , Fully Attempted. Item 3 , Unattempted. Item 4 , Unattempted. Item 5 , Unattempted.
To solve this problem, we can set up an exponential equation:
A = P(1 + r/n)^(n*t)
Where:
A = the ending balance
P = the initial deposit
r = the annual interest rate
n = the number of times interest is compounded per year
t = the number of years
In this case, we have:
A = $6,000
P = $5,000
r = 3% = 0.03
n = 1 (as the interest is compounded annually)
Plugging in these values into the equation, we have:
$6,000 = $5,000(1 + 0.03/1)^(1*t)
Simplifying the equation:
1.2 = (1.03)^t
Taking the logarithm of both sides to solve for t:
log(1.2) = log(1.03)^t
Using the logarithmic property log(a)^b = b * log(a):
log(1.2) = t * log(1.03)
Solving for t:
t = log(1.2) / log(1.03)
Using a calculator, we find:
t ≈ 9.617
Therefore, it will take approximately 10 years for the account balance to reach at least $6,000.
The correct exponential function to model the quail population is ƒ(x) = 175(1.22)x.
To find the approximate population after 5 years, we substitute x = 5 into the function:
ƒ(5) = 175(1.22)5
ƒ(5) ≈ 473.26
Therefore, the approximate population after 5 years is 473 quail.
Suppose you deposit $5,000 in a savings account that earns 3% annual interest. If you make no other withdrawals or deposits, how many years will it take the account balance to reach at least $6,000? (1 point) Responses 10 years 10 years 6 years 6 years 7 years 7 years 4 years
The population of wild horses on a wildlife refuge decreased from 110 to 99 in a year. If this trend were to continue, what exponential equation could be used to predict the horse population in the future? (1 point) Responses y = 110(1.1)x y = 110(1.1) x y = 99(0.9)x y = 99(0.9) x y = 99(1.1)x y = 99(1.1) x y = 110(0.9)x
To predict the horse population in the future using an exponential equation, we can use the formula:
y = a(1 + r)^x
Where:
y is the population in the future
a is the initial population
r is the growth rate per year
x is the number of years in the future
In this case, the initial population is 110 and it decreased to 99 in one year. Since the population is decreasing, the growth rate should be negative.
The correct exponential equation to predict the horse population would be:
y = 99(0.9)^x
This equation represents a decrease of 10% per year (-0.1) from the initial population of 99.
To write an exponential function to model the quail population, we can use the general form of an exponential function: f(x) = ab^x.
In this case, the initial population of quail is given as 175, and it increases at an annual rate of 22%. The growth rate is represented by the base of the exponential function, which is calculated by adding 1 to the given rate and converting it to a decimal.
So, the exponential function that models the quail population can be written as: f(x) = 175(1 + 0.22)^x.
Now, to find the approximate population after 5 years, we substitute x = 5 into the function.
f(5) = 175(1 + 0.22)^5
Calculating this, we get:
f(5) = 473
Therefore, the approximate population after 5 years is 473 quail.
So the correct answer is: ƒ(x) = 175(1.22)^x; 473.