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weeeeeb [17]
4 years ago
12

47 1/2% in simplest form

Mathematics
1 answer:
Leya [2.2K]4 years ago
5 0
The answer would be 47.5
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20% of 4hours 20 minutes
OverLord2011 [107]

Answer:

52 min

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for h
mezya [45]
So let's say the amounts are "a" and "b", at 6% and 8% respectively.

how much is 6% of a?  well (6/100) * a, or 0.06a.
how much is 8% of b?  well (8/100) * b, or 0.08b.

so hmm whatever "a" and "b" amounts are, we know that their yield is $100, thus 0.06a + 0.08b = 100.

now, there's more money in "a" than in "b", 500 bucks more, so whatever "b" is, we know that a = b + 500.

\bf \begin{cases}
0.06a+0.08b=100\\
\boxed{a}=b+500\\
----------\\
0.06\left( \boxed{b+500} \right)+0.08b=100
\end{cases}
\\\\\\
0.06b+0.08b+30=100\implies 0.02b=70\implies b=\cfrac{70}{0.02}
\\\\\\
b=3500

how much was deposited in the 6% one?  well, a = b + 500.
8 0
4 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
If one leg of a right triangle is 6 cm long, and the other leg is 8 cm long, how long is the triangles hypotenuse?!!
nekit [7.7K]

Answer:

approximately 9

Step-by-step explanation:

6 squared (36) plus 8 squared (64) is 84, and the square root of that is about 9

3 0
3 years ago
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Give an example of a set of five positive numbers that have a median of 10 and whose mean is larger than ten?
barxatty [35]
The answer is {1, 2, 10, 50, 75}. It has a median of 10 and a mean larger than 10
3 0
3 years ago
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