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3 years ago

Distribute 7 (2x2 – 2x).

Mathematics

1 answer:

Lisa [10]3 years ago

4 0

Answer:

If it is 2*2

Step-by-step explanation:

7(4-2x)

28-14x

I hope this helps!!

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How many numbers lie between the square of 56 and 57

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Given, two numbers 56 and 57 we need to find out the numbers lie between the squares of the given numbers.

Now, we have numbers lying between the square of n and (n + 1) is 2n

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3 years ago

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of flights i

Morgarella [4.7K]

Answer:

a) $P(X = n) = (0.85)^{n}$

b) $P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

d) If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrived on time, or it did not. Flights are independent. This means that 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A measure is said to unusual if it is more than 2.5 standard deviations from the mean.

85% of recent flights have arrived on time.

This means that $p = 0.85$

(a) Find the probability that all of the flights were on time.

This is $P(X = n)$ , in which n is the size of the sample studied.

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

$P(X = n) = C_{n,n}.(0.85)^{n}.(0.15)^{n-n}$

$P(X = n) = (0.85)^{n}$

(b) Find the probability that exactly x of the flights were on time.

Exactly x of the flights: So

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

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(c) Find the probability that x or more of the flights were on time.

This is

$P(X \geq x) = P(X = x) + P(X = x+1) + ... + P(X = n)$

For each value, we use

$P(X = x) = C_{n,x}.(0.85)^{x}.(0.15)^{n-x}$

(d) Would it be unusual for x or more of the flights to be on time?

If $x > E(X) + \sqrt{V(X)}$ , yes, otherwise no

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3 years ago