Respuesta :
For this case we have a function of the form:
[tex]y = A * b ^ x [/tex]
Where,
A: initial population
b: growth rate
x: time in hours
y: population after x hours
We must find the values of A and b, for this we use the following data:
After 1 hour, the population is 18:
[tex]18 = A * b ^ 1 [/tex]
After 2 hours, the population is 27:
[tex]27 = A * b ^ 2 [/tex]
We have a system of two equations with two unknowns
Dividing equations we have:
[tex] \frac{A * b ^ 2}{A * b ^ 1} = \frac{27}{18} [/tex]
[tex]b= \frac{27}{18} [/tex]
[tex]b = 1.5 [/tex]
Substituting b in equation 1 we have:
[tex]18 = A * (1.5) ^ 1 [/tex]
Clearing A we have:
[tex]A = 18 / 1.5 A = 12[/tex]
So, the function is:
[tex]y = 12 * (1.5) ^ x [/tex]
Answer:
A function that best describes the relationship between the time, in hours, and the population of the bacterium is:
C: EXPONENTIAL
the y-intercept of the function:
B: 12
the rate of change of the function:
C: MULTIPLY 1.5
[tex]y = A * b ^ x [/tex]
Where,
A: initial population
b: growth rate
x: time in hours
y: population after x hours
We must find the values of A and b, for this we use the following data:
After 1 hour, the population is 18:
[tex]18 = A * b ^ 1 [/tex]
After 2 hours, the population is 27:
[tex]27 = A * b ^ 2 [/tex]
We have a system of two equations with two unknowns
Dividing equations we have:
[tex] \frac{A * b ^ 2}{A * b ^ 1} = \frac{27}{18} [/tex]
[tex]b= \frac{27}{18} [/tex]
[tex]b = 1.5 [/tex]
Substituting b in equation 1 we have:
[tex]18 = A * (1.5) ^ 1 [/tex]
Clearing A we have:
[tex]A = 18 / 1.5 A = 12[/tex]
So, the function is:
[tex]y = 12 * (1.5) ^ x [/tex]
Answer:
A function that best describes the relationship between the time, in hours, and the population of the bacterium is:
C: EXPONENTIAL
the y-intercept of the function:
B: 12
the rate of change of the function:
C: MULTIPLY 1.5
Answer:
1. exponential 2. 12 3. multiply 1.5
Step-by-step explanation: