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Leya [2.2K]
3 years ago
11

I NEED HELP 17 POINTS PLUS BRAINLIST PLEASE

Mathematics
2 answers:
Sonja [21]3 years ago
4 0

Answer:

E

Step-by-step explanation:

Hope this helped!

KatRina [158]3 years ago
4 0
The answer is E because I did it and got it right
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Hi guys! I need the answer quick for 10 points! Thx!
svlad2 [7]

D . A cube that is 1 inch long , 1inch wide and 1 inch high .

\red{ \bold{What \: is \: unit \: cube \: }}

Answer

its a cube whose side is 1 unit long .

And total surface area should be 6 square units .

as we can see

3 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

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
Va rog,e urgen!!!!!
coldgirl [10]
5 is the answer I think
8 0
3 years ago
greta has a piece of cloth that is 9 yards long . she cuts it into pieces that are each 1/3 yards long how many pieces of cloth
labwork [276]
27 peices i believe. 3 peices per yard, 9 yards. 9*3=27
7 0
1 year ago
PLEASE HELP WILL MARK BRAINLIEST!
Naily [24]

Answer:

The answer is given in the picture

Please mark me brainliest

7 0
2 years ago
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