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3 years ago

7

URGENT PLEASE HELP!!! **will mark brainliest

1 answer:

AveGali [126]3 years ago

8 0

The answer is option B. The real answer is 8.5, not 0.85

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Please help me out here! Any of these questions!

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A line segment ad, contains points b & c such that c is between a and d, and b is between a and

egoroff_w [7]

If you drew out this line segment with the points in the order we are given them, the segment would be labeled as a, b, c, d in that order. We are given a measure for ab, and we are also given a measure for bd with c being ignored for a minute. The entire length of the segment can be found by adding ab and bd. 6 + 23 = ad and ad = 29. So the length of the whole segment is 29. We have the length of cd given to be equal to ab, so cd = 6. ab + bc + cd = ad and we are looking for bc. Since we have the other lengths and the length of the whole segment, we fill in accordingly: 6 + bc + 6 = 29. Solving for bc we get bc = 17.

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3 years ago

Daily income of 8 workers is Rs. 2400. How many workers will earn Rs.1800

ValentinkaMS [17]

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Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used 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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

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

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

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4 years ago