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Neporo4naja [7]
2 years ago
14

PLZ HELP, WILL GIVE BRAINLIEST!!!

Mathematics
1 answer:
ahrayia [7]2 years ago
7 0

Answer:

i would say a senior citizen ticket i 3 dollars and the child tickets are 5 dollars

Step-by-step explanation:

You might be interested in
Can someone help me with this-
kodGreya [7K]

we can take the recipricol

so  

(6/7)x2

=12/7

hope this helps :DDDD

brainliest plsss

5 0
3 years ago
Read 2 more answers
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
Daniel went to Target and bought a basketball for $ 21.85 and a shirt for $12.68. Daniel received $ 0.47 in change. How much did
jarptica [38.1K]

Answer:

$35

Step-by-step explanation:

To do this you would just add the 2 prices together and you would get $34.53 and you would add 47 cents to that and you would get that Daniel gave $35 to the cashier.

5 0
3 years ago
I need help if y’all want y’all can help me I need help
Snezhnost [94]

The answer is D. 42

A triangle equals 180 in total so what you would do it subtract 122 from 180 because 180 is also the angle of a straight line. You would get 58 then you add 58 and 80 to get 138. Subtract 138 from 180 and you get your answer of 42

6 0
3 years ago
What is the mean (average) of these five numbers?
IRINA_888 [86]

Answer:

C. 5

Step-by-step explanation:

Mean: Numbers over amount of numbers

(2 + 4 + 5 + 6 + 8)/5

25/5

5

4 0
2 years ago
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