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2 years ago

The solution set of an absolute value inequality is shown.Please no links

Mathematics

1 answer:

larisa86 [58]2 years ago

5 0

Answer:

x≤ -4 ≤ 2

Step-by-step explanation:

I am honestly not sure but this seems right. Good luck.

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After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the prob

Westkost [7]

Answer:

a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

Step-by-step explanation:

For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. So 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.

15% of all buildings have been structurally compromised.

This means that $p = 0.15$

20 buildings

This means that $n = 20$

a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

This is P(X = 1).

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

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756$

17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702$

17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028$

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570$

75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

The expected value of the binomial distribution is:

$E(X) = np$

$E(X) = 20*0.15 = 3$

The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

3 0

2 years ago

May B. Luckie tosses a dart at the April calendar on the wall. What are the ODDS that the dart hits a date that is a prime numbe

meriva

Answer:

The ODDS that the dart hits a date that is a prime number are 1:1

Step-by-step explanation:

Odds:

An odd is writen as:

Number of desired outcomes: Number of undesired outcomes.

In this question:

April has 30 days, 15 of which are prime numbers and 15 of which are odd numbers.

So the odds are:

15:15

Simplifying by 15

1:1

The ODDS that the dart hits a date that is a prime number are 1:1

5 0

2 years ago

In January of 2011, the U.S. saw an increase in gas prices. Imagine the average price per gallon was $3.12 with a standard devia

Arada [10]

Answer:

31.92%

Step-by-step explanation:

We are given;

Population mean; μ = $3.12

Sample mean; x¯ = $3.16

Sample size; n = 10

Standard deviation; σ = $0.27

Z-score formula is; z = (x¯ - μ)/(σ/√n)

z = (3.16 - 3.12)/(0.27/√10)

z = 0.04/(0.08538)

z ≈ 0.47

Now, the percent of other sample means, based on 10 gas stations, that would be greater than the one observed is;

P(x¯ > 3.12) = 1 - P(z < 0.47)

From z-table attached P(z < 0.47) = 0.68082

Thus;

P(z > 0.47) = 1 - 0.68082

P(z > 0.47) ≈ 0.3192

This expressed in percentage is 31.92%

4 0

2 years ago

Can you please help me?!

Fudgin [204]

Answer:

A and B are correct

Step-by-step explanation:

4 0

2 years ago

Find the midpoint of A and B where A has coordinates (2, 4)<br> and B has coordinates (-3, -9).

tatuchka [14]

Answer:

(-0.5,-2.5)

Step-by-step explanation:

(x1 + x2) / 2 = x midpoint
(y1 + y2) / 2 = y midpoint

⇒ 2 + -3 = 5

⇒ 5 / 2 = -0.5

⇒ 4 + -9 = -5

⇒ -5 / 2 = -2.5

→ = (-0.5, -2.5)

4 0

2 years ago

Read 2 more answers

The solution set of an absolute value inequality is shown.Please no links​

The solution set of an absolute value inequality is shown.Please no links