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pogonyaev
3 years ago
12

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume

that passengers average 196 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 32 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 18 passengers. What is the approximate probability that the total weight of the passengers exceeds 3928 pounds
Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer: 0.0016

Step-by-step explanation:

Given the following :

Population mean(m) = 196

Population Standard deviation (sd) = 32

Number of sample (n) = 18

Sample mean (x) = Total weight of sample / number of samples = 3928 / 18 = 218.2

Obtaining the test statistic (z) :

Z = [(x - m) / (sd/√n)]

Z = [(218.2 - 196) / (32/√18)]

Z = [(22.2 / 7.5424723)]

Z = 2.943

Hence,

P(Z > 2.94) = 1 - P(Z < 2.94)

= 1 - 0.9984

= 0.0016

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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
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If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

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So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

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To know more about factorization, refer: brainly.com/question/25829061

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