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3 years ago

The sum of the two prime numbers between 110 and 130 is

1 answer:

7 0

Prime numbers are those numbers which have only two factors, 1 and the number itself . For e.g. 3,5,7,11 etc are all prime numbers.

Between 110 and 130 we have a total of two prime numbers which are 113 and 127, since both have only two factors .

And sum of 113 and 127 is

$113+127= 240$

So the required answer is 240 .

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Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

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Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

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.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

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3 years ago

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Jet001 [13]

The quadratic formula is above.

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I begin each problem by defining variables.
For instance 3. Is in standard form. X^2 -2x - 3 = 0. a = 1, b = -2, c = -3

Now use quadratic formula: x = [- b + or - sqrt(b^2 - 4ac)]\2a

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3 years ago