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3 years ago

Please Please Please Help Me

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

(x-y^2)(x^2 +xy^2 +y^4)
(a^2 +b)(a^4 -a^2b +b^2)
(m^3-n)(m^6 +m^3n +n^2)
(p+k^3)(p^2 -pk^3 +k^6)
(a^2+b^3)(a^4 -a^2b^3 +b^6)
(x-y)(x^2 +xy +y^2)(x^6 +x^3y^3 +y^6)

Step-by-step explanation:

In every case, the factorization makes use of the standard form for factoring the sum or difference of cubes:

a^3 +b^3 = (a +b)(a^2 -ab +b^2)
a^3 -b^3 = (a -b)(a^2 +ab +b^2)

1. a=x, b=y^2. Use the formula for the difference.

2. a^2 ⇒ a, b = b. Use the formula for the sum.

3. a=m^3, b=n. Use the formula for the difference.

4. a=p b=k^3. Use the formula for the sum.

5. a^2 ⇒ a, b^3 ⇒ b. Use the formula for the sum.

6. a=x^3, b=y^3. Use the formula for the difference. When you do, the first factor is the difference x^3 -y^3, which can be factored using the difference formula again with a=x, b=y.

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Divide 0.008 divided by 0.25.

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20 points Return to questionItem 4Item 4 20 points Police records in the town of Saratoga show that 13 percent of the drivers st

Sladkaya [172]

Answer:

a) 0.1423

b) 0.2977

c) 0.56

Step-by-step explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that $p = 0.13$

14 drivers are stopped

This means that $n = 14$

(a) None will have an invalid license.

This is $P(X = 0)$

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

$P(X = 0) = C_{14,0}.(0.13)^{0}.(0.87)^{14} = 0.1423$

(b) Exactly one will have an invalid license.

This is $P(X = 1)$

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

$P(X = 1) = C_{14,1}.(0.13)^{1}.(0.87)^{13} = 0.2977$

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56$

8 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

2 years ago