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dem82 [27]
3 years ago
8

A fish population, p, can be represented by the equation p = 800 . 1/2 where t is time in years since the beginning of 2015. Wha

t was the fish population at the beginning of 2012? A. 100 B. 800 C. 2,400 D. 6,400
Mathematics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

I believe its A but I am not 100% positive

Step-by-step explanation:

You might be interested in
What percent of 18 is 24? Side info: the answer is not 75% or 133%!!!
vredina [299]

Answer:

133 1/3 %

Step-by-step explanation:

24/18 = 4/3

4/3 * 100 = 133 1/3 %

Missed that 1/3

3 0
4 years ago
Read 2 more answers
(08.01 MC)
Colt1911 [192]

Part A:

We see that this pair of equations has 1 solution. On a graph, the solutions of 2 (or more) lines is where the lines intersect. In this case, since these lines intersect 1 time, they have 1 solution.

Part B:

As previously mentioned, the solutions of multiple lines is where the lines intersect. In this case, since they intersect at (4,4), that is the solution.

6 0
3 years ago
Does anyone know how to do this/have the answer? PLEASE
podryga [215]

15: You solve problems like this by finding the probability of each case, and then multiplying them all. For each of the last 4 question, she has probability 1/2 of guessing right. So, she guessed 4 consecutive questions with probability

\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}

16: Like before: you have pick a king with probability 4/52 = 1/13 (there are four kings - one for each suit, out of 52 cards in a standard deck), and you pick "I" from "INCREDIBLE" with probability 2/10 = 1/5 (there are two "I"s out of 10 letters). So, the probability of picking a king and then an "I" is

\dfrac{1}{13}\cdot\dfrac{1}{5}=\dfrac{1}{65}

17-20: The important bit of information here is that you replace the first ball. So, the first and second pick follow the exact same probability distribution, because they basically are two repetitions of the same experiment. So, for example, in ex. 17, the first ball is even with probability 15/30 = 1/2 (there are 15 even balls out of 30). Then, you have again probability 15/30 = 1/2 to pick an odd ball (there are also 15 odd balls out of 30). So, the probability of picking an even ball, replace it, and pick an odd ball is

\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}

Exercises 18 to 20 follow the same scheme: find out the probability of the two events and multiply them.

21-26: Not we DON'T replace the balls, so the second pick will suffer the effects of the first one. Let's dive into ex. 21 for example. For the first pick, we want a 2-digits number. There are 21 of such balls (all balls except balls 1 to 9), so we pick a 2-digits ball with probability 21/30 = 7/10. For the second pick, we want the balls number 4. But we have to assume that we already picked the first ball, and we picked a 2-digits ball. So, for the second pick, we're choosing from a bag with 29 balls, and there is only one ball labeled 4. So, we pick the ball number 4 with probability 1/29. We deduce that the two events happen one after the other with probability

\dfrac{7}{10}\cdot\dfrac{1}{29}=\dfrac{7}{290}

Exercises 22 to 26 are similar: you find out the probability of the first event, and then you consider the new environment (i.e. you keep track of the first pick) when it comes to the probability distribution for the second pick).

5 0
3 years ago
Pls help help help help​
Thepotemich [5.8K]

Answer:

D

Step-by-step explanation:

We can plug in the numbers (15, 23, 25, 38, 53) into the equation for x, and see if we get the values given for the number of hits (4, 12, 14, 27, 47)

8 0
3 years ago
Trong mặt phẳngOxy , phép quay mỗi
NeTakaya

Answer:

Step-by-step explanation:

4 0
3 years ago
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