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3 years ago

Which statement best describes how to determine whether f(x) = x3 + 5x + 1 is an even function?

1 answer:

7 0

I have my notes here that will help you in answering the problem:

Test to determine if a function y=f(x) is even, odd or neither: Replace x with -x and compare the result to f(x). If f(-x) = f(x), the function is even. If f(-x) = - f(x), the functionis odd.

I hope my answer has come to your help. God bless and have a nice day ahead!

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HELP! Need quick - Determine whether the data distribution appears to be positively skewed, negatively skewed, or symmetric.

zalisa [80]

Answer:

A. negatively skewed

Step-by-step explanation:

Plot the values of salary against years

You will notice that;

The graph has a long left tail
The tail is longer towards the negative direction of the number line

This mean that this graph is a negative skewed graph/left-skewed distribution

In the attached sketch of the plot, join the points with a smooth curve and observe the above mentioned properties.

The x-axis scale ranges from 2000 to 2015 where 0=2000, 1=2001,2=2002......,15=2015

Hope this will give you a visual picture of the negatively- skewed distribution

8 0

3 years ago

Bonnie is making a dipping sauce. She mixes 150 milliliters of soy sauce with 100 milliliters of vinegar. How much soy sauce doe

natali 33 [55]

First step

You must stablish a relation

150 ml sauce/100 ml vinegar

If 100 ml vinegar_______150ml sauce then 11 ml vinegar ____x ml sauce

Second step

You must resolve x from the equation

X= (11ml vinegar*150 ml soya)/(100 ml vinegar)= 16.5 ml sauce, it means that it will be 16,5 ml sauce by every 11 ml of vinegar

5 0

4 years ago

Read 2 more answers

A. y=8x+0 b. x=0 c. y=0x +8 d. x=8 PLEASE ANSWER ASAP!!

ki77a [65]

Answer:

option d. x = 8

Step-by-step explanation:

8 0

3 years ago

Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0

4 years ago

3. The following are the scores you shot in five rounds of golf: 70,80,75,75,80. The variance is: A. 3.74 C. 14 B. 196 D. 70

Paladinen [302]

Answer:

Option C.

Step-by-step explanation:

For a given set of N elements:

{x₁, x₂, ..., xₙ}

The mean is given by:

$M = \frac{x_1 + x_2 + ... + x_n}{N}$

And the variance is given by:

$v = \frac{(x_1 - M)^2 + (x_2 - M)^2 + ... + (x_n - M)^2}{N}$

Then for our set:

{70, 80 ,75, 75, 80}

We have 5 elements, then N = 5

The first thing we need to find is the mean.

$M = \frac{70 + 80 + 75 + 75 + 80}{5} = 76$

Then the variance will be:

$v = \frac{(70 - 76)^2 + (80 - 76)^2 + (75 - 76)^2 + (75 - 76)^2 + (80 - 76)^2}{5} = 14$

The correct option is C, 14.

6 0

3 years ago