Can someone plz help me wit this?

Answer:

There should be at most 24 lucky numbers in the third bag.

Step-by-step explanation:

Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.

Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:

200 - 49

300 - x

With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.

Answer:

(x - 1) (x - 2) (x + 2) (x + 3)

Step-by-step explanation:

Factor the following:

x^4 + 2 x^3 - 7 x^2 - 8 x + 12

The possible rational roots of x^4 + 2 x^3 - 7 x^2 - 8 x + 12 are x = ± 1, x = ± 2, x = ± 3, x = ± 4, x = ± 6, x = ± 12. Of these, x = 1, x = 2, x = -2 and x = -3 are roots. This gives x - 1, x - 2, x + 2 and x + 3 as all factors:

Answer: (x - 1) (x - 2) (x + 2) (x + 3)

Answer:

The probability that he or she is high-risk is 0.50

Step-by-step explanation:

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10  

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to  calculate P( High risk | at-fault accident) = ?

Using Bayes' conditional probability theorem

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) /  { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) +  P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50.

Final answer:

The probability that a driver is high-risk given that they had an at-fault accident can be found using Bayes' theorem. Given the probabilities provided in the question, the probability is approximately 0.3333 or 33.33%.

Explanation:

To find the probability that a driver is high-risk given that they had an at-fault accident, we can use Bayes' theorem. Let's define the events:

1. A: Driver is high-risk
2. B: Driver has an at-fault accident

We are given the following probabilities:

1. P(A) = 0.20 (probability of a driver being high-risk)
2. P(B|A) = 0.20 (probability of an at-fault accident given that they are high-risk)
3. P(~A) = 0.80 (probability of a driver not being high-risk)
4. P(B|~A) = 0.10 (probability of an at-fault accident given that they are not high-risk)

Using Bayes' theorem, the probability of a driver being high-risk given that they had an at-fault accident is:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(~A) * P(B|~A))

Substituting the given probabilities:

P(A|B) = (0.20 * 0.20) / (0.20 * 0.20 + 0.80 * 0.10) = 0.04 / (0.04 + 0.08) = 0.04 / 0.12 = 0.3333.

Therefore, the probability that a driver is high-risk given that they had an at-fault accident in the next year is approximately 0.3333 or 33.33%.

Learn more about probability here:

brainly.com/question/32117953

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Answer:

The answer is

Step-by-step explanation:

we know that

A quotient is the division of two numbers

In the expression divided by n

the numerator is equal to

the denominator is equal to

so

the quotient is equal to

The complete expression divided by n plus is equal to