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kakasveta [241]
2 years ago
9

A line has a slope of 6 and a y-intercept of 5. What is its equation in slope-intercept form?

Mathematics
1 answer:
kvasek [131]2 years ago
3 0

Answer:

The equation in slope-intercept form is y = 6x + 5

Step-by-step explanation:

The slope-intercept form of the linear equation is y = m x + b, where

  • m is the slope
  • b is the y-intercept

Let us solve the question

∵ A line has a slope of 6

∵ m is the slope of the line

∴ m = 6

∵ The line has a y-intercept of 5

∵ b is the y-intercept

∴ b = 5

→ Substitute the values of m and b in the slope-intercept form

∵ y = m x + b

∴ y = 6x + 5

∴ The equation in slope-intercept form is y = 6x + 5

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Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s
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Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that 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.

No-show rate of 10%.

This means that p = 0.1

Four travelers who have made hotel reservations in this study.

This means that n = 4

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

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

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523

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b) What is the most likely value for X?

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

P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561

P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916

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P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

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THIS IS FOR 27 POINTS!!!<br> EXPLAIN PLEASE!!!!<br> Solve -9(2+a) + 4(2a+9)=11
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---------------------
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Divide both sides by -1
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