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luda_lava [24]
3 years ago
11

47% of citizens in a district voted in the last election. 21% of citizens in the district are under 18. Being younger than 18 an

d voting in the last election are mutually exclusive. For a randomly selected citizen in the district, answer the following: What is the probability the citizen voted in the last election or was under 18?
Mathematics
1 answer:
Darya [45]3 years ago
4 0

Answer:

68% probability the citizen voted in the last election or was under 18.

Step-by-step explanation:

Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.

So, the probability of a citizen voting in the last election AND being under 18 is 0%.

What is the probability the citizen voted in the last election or was under 18?

Since these events are mutually exclusive, this probability is the sum of each separate probability.

So there is a 47+21 = 68% probability the citizen voted in the last election or was under 18.

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If you rolled two dice what is the probability that you would roll a 4 or a 5
Masja [62]
Hello there! Given that normal dice are numbered 1-6, individually rolling a 4 or a 5 would give a 1/6 probability. That converts to about 17% as a percentage because we can multiply 1 by 100 to get 100/6, then divide 100 by 6 to get 16.6666. When rounding, that gives approximately 17%. However, if we combined probabilities, we would find that rolling a 4 or a 5 collectively gives a 2/6 probability, which is approximately 33% as a decimal.

In terms of individual probabilities, you would be 17% likely to roll one of them. In terms of collectiveness, the likelihood of rolling a 4 or 5 would render 33% on each die. If you need additional help, let me know and I will gladly assist you.
5 0
2 years ago
The marginal profit in dollars on Brie cheese sold at a cheese store is given by Upper P prime (x )equals x (60 x squared plus 9
blondinia [14]

Answer:

Part A:

P(x)=15x^4+30x^3-50

Part B:

P(4)=$4270

Step-by-step explanation:

Part A:

In order to find the profit function P(x) we have to integrate the P'(x)

P'(x)=x(60x^2+90x)

P'(x)=60x^3+90x^2

\int\limits {p'(x)} \, dx =\int\limits {60x^3+90x^2} \, dx

P(x)=15x^4+30x^3+C

when x=0, C=-50

P(x)=15x^4+30x^3-50

Part B:

x=4

P(x)=15x^4+30x^3-50

P(4)=15*4^4+30*4^3-50

P(4)=$4270

Profit from selling 400 pounds is $4270

8 0
3 years ago
Math question plz help
Svetach [21]

Answer:

Yes they are.

Because:

As said in the rules you have to keep the bases and add the exponents.

7 0
3 years ago
Read 2 more answers
Since people are sad they missed my 50 points, here is 100 points for the first 2 to answer. First one gets branliest aswell.
svetoff [14.1K]
Ummm is there something I Guccie
6 0
3 years ago
The points shown on the graph represent the numbers in a geometric sequence. What is the initial value of the sequence? 1 2 3 8
Lina20 [59]

Answer:

the answer is 3 just took the test

Step-by-step explanation:

8 0
3 years ago
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