I really think the problem I labeled 1 is the correct interpretation but just in case you meant the latter I wrote the latter as well. The sentence translates to exactly what I have for problem number 1.

The Problem:

1) Determine if $f(x)=x+\frac{4}{x}$ is even, odd, or neither.

2) Determine if $f(x)=\frac{x+4}{x}$ is even, odd, or neither.

Step-by-step explanation:

$f(-x)=f(x)$ implies $f$ is even.

$f(-x)=-f(x)$ implies $f$ is odd.

So either definition says we have to plug in $-x$ .

$f(x)=x+\frac{4}{x}$ with new input $-x$ :

$f(-x)=-x+\frac{4}{-x}$

$f(-x)=-x+-\frac{4}{x}$

$f(-x)=-(x+\frac{4}{x})$

$f(-x)=-f(x)$

This means $f$ is even since we got the same thing we started with.

$f(x)=\frac{x+4}{x}$ with new input $-x$ :

$f(-x)=\frac{-x+4}{-x}$

$f(-x)=\frac{-(x-4)}{-x}$

$f(-x)=\frac{x-4}{x}$

This is neither the same or the opposite of what we started with.

5 0

4 years ago

An airliner carries 150 passengers and has doors with a height of 78 in. Heights of men are normally distributed with a mean of

iogann1982 [59]

Answer:

Step-by-step explanation:

x, height of men is N(69, 2.8)

Sample size n =150

Hence sample std dev = $\frac{2.8}{\sqrt{150} } =0.229$

Hence Z score = $\frac{x-69}{0.229}$

A) Prob that a random man from 150 can fit without bending

= P(X<78) = P(Z<3.214)=1.0000

B) n =75

Sample std dev = $\frac{2.8}{\sqrt{75} } =0.3233$

P(X bar <72) = P(Z<9.28) = 1.00

C) Prob of B is more relevent because average male passengers would be more relevant than a single person

(D) The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.

3 0

3 years ago