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3 years ago

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Nathan has a rectangular plot of grass in his backyard that has dimensions of meters in width and meters in length. What is the

total area of the plot of grass? Recall that the area of a rectangle is the width multiplied by the length. square meters square meters square meters square meters

2 answers:

trapecia [35]3 years ago

5 0

Answer:

B on edge

Step-by-step explanation:

Hope it helps!

jek_recluse [69]3 years ago

4 0

Answer:

B

Step-by-step explanation:

took quiz

You might be interested in

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:

$E(X) = np$

So

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

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

There are 3 meters from A to B, 22 meters from B to C, 5 meters from C to D, and 7 meters from D to E. What is the distance from

notka56 [123]

17m

Because horizontal distance is 8m, vertical distance is 15m
So distance between A and Enid square root of 8^2+15^2

8 0

3 years ago

There are about 448 students fully virtual and about 92 students coming into school for hybrid learning. Write and simplify a ra

lutik1710 [3]

448: 92
Divide by 4
112 : 23

4 0

2 years ago

Read 2 more answers

It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportat

Mrrafil [7]

Answer:

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 120$

Alternative hypothesis: $\mu \neq 120$

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z$

Step-by-step explanation:

Data given and notation

$\bar X=114$ represent the sample mean

$\sigma=22$ represent the population standard deviation

$n=36$ sample size

$\mu_o =120$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part 1

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 120$

Alternative hypothesis: $\mu \neq 120$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z$

8 0

3 years ago

Plz someone help me on this too plz plz

san4es73 [151]

1. 7 and -7 are both 7 units away from 0. They are 14 units away from each other. Absolute value helps because rather than 7+(-7) being 0 absolute value makes anything positive so it becomes 7+7=14 to be able to count the units between the two.

2. 5 and -5 are both 5 units away from 0. They are 10 units away from each other. Absolute value helps because rather than 5+(-5) being 0 absolute value makes anything positive so it becomes 5+5=10 to be able to count the units between the two.

3. 2 and -2 are both 2 units away from 0. They are 4 units away from each other. Absolute value helps because rather than 2+(-2) being 0 absolute value makes anything positive so it becomes 2+2=4 to be able to count the units between the two.

Hope that helps

8 0

3 years ago