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3 years ago

PLEASE HELP!!! ILL GIVE BRAINLIEST

Mathematics

2 answers:

MakcuM [25]3 years ago

7 0

Answer:

*gives help*

Step-by-step explanation:

weeeeeb [17]3 years ago

5 0

I’m sorry I live in Canada- Toronto

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I WILL MARK AS BRAINLIEST!!!A man selling hot dogs, french fries, and a soda as a combo meal at a fairground has to pay the coun

9966 [12]

Answer:

Step-by-step explanation:

I think the answer is C, I am not 100% sure tho

5 0

3 years ago

If K = (AB)/(A+B) , then B = ?

marysya [2.9K]

Lets do

$\\ \sf\longmapsto K=\dfrac{AB}{A+B}$

$\\ \sf\longmapsto \dfrac{1}{K}=\dfrac{A+B}{AB}$

$\\ \sf\longmapsto \dfrac{1}{K}=\dfrac{1}{A}+\dfrac{1}{B}$

$\\ \sf\longmapsto \dfrac{1}{B}=\dfrac{1}{K}-\dfrac{1}{A}$

$\\ \sf\longmapsto \dfrac{1}{B}=\dfrac{K-A}{AK}$

$\\ \sf\longmapsto B=\dfrac{AK}{K-A}$

6 0

3 years ago

Find the probability that when 200 reservations are accepted for united flight 15, there are more passengers than seats availabl

Arlecino [84]

The solution for this problem is:

We know the problem has the following given:

Sample size of 200

X = 182

And the probability of .9005; computation: 1 - .0995 = .9005

So in order to get the probability:

P (x >= 182) = 1 – 0.707134 = .292866 is the probability that when 200 reservations
are recognized, there are more passengers showing up than there
are seats vacant.

The other solution is:

p (>= 182) = p(183) + P(184) + P(185) + ... + P(199) + P(200) = 0.292866

7 0

4 years ago

Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and

Olenka [21]

Answer:

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

$P(A \cap B) = P(A)*P(B)$

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)$

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

$P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)$

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}$

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

$P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}$

Probability that the series lasts exactly four games:

$p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})$

The probability that the series lasts exactly four games is $3p(1-p)(p^{2} + (1 - p)^{2})$

8 0

3 years ago

The value of 2^3 + 3^3 = ___. Numerical Answers Expected!

blsea [12.9K]

Solve the exponents first:

$2^{3} = 2 \times 2 \times 2 = 8$

$3^{3} = 3 \times 3 \times 3 = 27$

Add the results together:

$8 + 27 = 35$

The answer is 35.

7 0

3 years ago

Read 2 more answers