Given:
Initial number of bacteria = 3000
With a growth constant (k) of 2.8 per hour.
To find:
The number of hours it will take to be 15,000 bacteria.
Solution:
Let P(t) be the number of bacteria after t number of hours.
The exponential growth model (continuously) is:

Where,
is the initial value, k is the growth constant and t is the number of years.
Putting
in the above formula, we get



Taking ln on both sides, we get

![[\because \ln e^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20e%5Ex%3Dx%5D)



Therefore, the number of bacteria will be 15,000 after 0.575 hours.
Answer:
$70,201.38
Step-by-step explanation:
To determine the principal amount, we can use the formula:

A = $200,000
r = 7% or 0.07
t = 15 years
n = 12
Now let's substitute our values.



So the principal needed to get 200,000 in 15 years is $70,201.38.
Option C
For each value of y, -2 is a solution of -21 = 6y - 9
<u>Solution:</u>
Given, equation is – 21 = 6y – 9
We have to find that whether given set of options can satisfy the above equation or not
Now, let us check one by one option
<em><u>Option A) </u></em>
Given option is -5
Let us substitute -5 in given equation
- 21 = 6(-5) – 9
- 21 = -30 – 9
- 21 = - 39
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option B)</u></em>
Given option is 3
- 21 = 6(3) – 9
- 21 = 18 – 9
- 21 = 9
L.H.S ≠ R.H.S ⇒ not a solution
<em><u>Option C)</u></em>
Given option is -2
- 21 = 6(-2) – 9
- 21 = - 12 – 9
- 21 = - 21
L.H.S = R.H.S ⇒ yes a solution
<em><u>Option D)</u></em>
- 21 = 6(9) – 9
- 21 = 54 – 9
- 21 = 45
L.H.S ≠ R.H.S ⇒ not a solution
Hence, the solution for the given equation is – 2, so option c is correct