All categories

3 years ago

Solve for x Pls help I need this for study guide

Mathematics

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

your study guide can be stuck up you know where :3

You might be interested in

What is 709032-249136?

alex41 [277]

709,032 - 239,136 = 469,896

Hope I Helped!!!

7 0

3 years ago

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

3 years ago

Write an equation of the parabola in intercept form with x-intercepts of 12 and -6; and an axis of symmetry of (14, 4).

xenn [34]

Intercept form is: y = a(x - p)(x - q)

It is given that: p = 14, q = -6, x = 14, y = 4

4 = a(14 - 12)(14 - (-6))

4 = a(2)(20)

4 = 40a

$\frac{4}{40} = \frac{40a}{40}$

$\frac{1}{10} = a$

Answer: y = $\frac{1}{10}$ (x - 14)(x + 6)

6 0

3 years ago

A bag with 8 marbles is shown below. (2 marbles are yellow, 3 are red, and 3 are blue.) A marble is chosen from the bag at rando

Andreas93 [3]

Answer:

$\frac{5}{8}$

Step-by-step explanation:

Hi there!

We're given that there are 8 marbles in total. 2 marbles are yellow and 3 are red, making the number of red AND yellow marbles 5 (2+3=5)

Because 5 out of the 8 marbles are yellow or red, the probability of choosing one of them is $\frac{5}{8}$ .

I hope this helps!

7 0

3 years ago

Read 2 more answers

The quotient of a number y and 2. A. 2/y B. 2(y) C. 2 + y D. y/2

Virty [35]

Answer:

The answer to your problem is B.

7 0

2 years ago