Answer:
510 gorillas
Step-by-step explanation:
In this problem, the population of gorillas is decreasing at a rate of 3.5 % per year.
We can write an expression for the population of gorillas as follows:
![n(t) = n_0 (1+r)^t](https://tex.z-dn.net/?f=n%28t%29%20%3D%20n_0%20%281%2Br%29%5Et)
where
n(t) is the number of gorillas after t years
is the number of gorillas at t = 0
r is the grow rate of the population
t is the number of years
Here we have:
is the number of gorillas when t = 20 years
is the grow rate of the population
So the equation becomes:
![250 = n_0 (1-0.035)^{20} = n_0 (0.965)^{20}](https://tex.z-dn.net/?f=250%20%3D%20n_0%20%281-0.035%29%5E%7B20%7D%20%3D%20n_0%20%280.965%29%5E%7B20%7D)
And solving for
, we find the initial number of gorillas:
![250=n_0 (0.965)^{20}=n_0 \cdot 0.490\\n_0 = \frac{250}{0.490}=510](https://tex.z-dn.net/?f=250%3Dn_0%20%280.965%29%5E%7B20%7D%3Dn_0%20%5Ccdot%200.490%5C%5Cn_0%20%3D%20%5Cfrac%7B250%7D%7B0.490%7D%3D510)
Answer:
Step-by-step explanation:
Evidently the quesiton is incomplete.
It is logic that the actions are roll a die and draw a card.
The two events are indepent, so the combined probability is the sum of the individual probabiliites.
Probability = number of positive events / number of possible events
Probability of rolling an even number = number of faces with even numbers / total number of numbers.
Probability of rolling an even number = 3 / 6 = 1/2
Probability of drawing an ace = number of aces in the deck of cards / number of cards
Probability of drawing an ace = 4 / 52 = 1 / 13
Combined probability = 1/2 + 1/13 = (13 + 2) / 26 = 15 / 26
Answer: option D. 15 / 26