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3 years ago

13

Liability insurance covers bodily injury to others and any property damaged, including yours.

2 answers:

ehidna [41]3 years ago

4 0

False, it only covers the other persons damages. That's what makes you "liable"

SSSSS [86.1K]3 years ago

3 0

Liability insurance covers bodily injury to others and any property damaged, including yours. This statement is TRUE.

There are several types of liability insurance. Auto liability insurance covers the damages that you've created to someone else's car and bodily injury.

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6. What does |-9| equal?

hichkok12 [17]

The absolute value makes it 9

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3 years ago

Read 2 more answers

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

notsponge [240]

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

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

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

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

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

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

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

b) What is the most likely value for X?

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

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

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

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

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

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

X = 0 has the highest probability, which means that 0 is the most likely value for X.

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2 years ago

Solving perimeter and area problems

arlik [135]

For perimeter you add all of the sides but for area you multiply each side

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2 years ago

I'm sorry but I have another one!

SIZIF [17.4K]

No mode or 43 i know the 43 sounds crazy but with math now days you never know.

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3 years ago

Please help! I mark brainliest :3

vladimir1956 [14]

no for a
yes for B
no for c
that what I think It is opinion could be wrong

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2 years ago