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3 years ago

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Prove that the sum of two odd functions is odd.

1 answer:

katrin [286]3 years ago

7 0

Odd function: f(-x) = -f(x)

Let two odd functions be f(x) and g(x)

Let k(x) = f(x) + g(x)

Then k(-x) = f(-x) + g(-x) = -(f(x) + g(x)) = -k(x)

So k(x) is odd too.

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What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phon

vovangra [49]

Answer:

Hence, the probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7887

Step-by-step explanation:

We are given the following information in the question:

We treat people using phones in the store for guidance as a success.

P(People use phone for guidance) = 57% = 0.57

Then the number of people follows a binomial distribution, where

Formula:

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 14 and x = 7

We have to evaluate:

$P(x \geq 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x =11) + P(x = 12) + P(x = 13) + P(x = 14)\\= \binom{14}{7}(0.57)^7(1-0.57)^7 + \binom{14}{8}(0.57)^8(1-0.57)^6 + \binom{14}{7}(0.57)^9(1-0.57)^5 + \binom{14}{7}(0.57)^{10}(1-0.57)^4 + \binom{14}{7}(0.57)^{11}(1-0.79)^3 + \binom{14}{7}(0.57)^{12}(1-0.57)^2 + \binom{14}{7}(0.57)^{13}(1-0.57)^1 + \binom{14}{7}(0.57)^{14}(1-0.57)^0\\= 0.7887 = 78.67\%$

Hence, the probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7887

5 0

3 years ago

Yy . tgt<br> ########$$$$$$$%$$$

Schach [20]

Answer:

yy . tgt

########$$$$$$$%$$$

4 0

2 years ago

Read 2 more answers

Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

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

or

⇒ $=\sum e^{tx}p(x)$

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

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,

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

or

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

also,

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

Thus,

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

or

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

Therefore,

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

or

Mean, E(X) = - 0.5

and

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

or

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

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

or

⇒ $=\sum e^{tx}p(x)$

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

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

or

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

If you don’t know how to solve this equation, please skip my question and answer someone else’s question(s). Using either slope-

Vsevolod [243]

Answer:

You were close; you don’t have to add 3x to the right side though.

We’re given 2y= -3x + 6

Step one: divide both sides by 2

y= (-3/2)x + 3

5 0

3 years ago

HELP ASAP <br><br> Convert the measurement.<br> 15 meters =___centimeters

BARSIC [14]

Answer:

1500

Step-by-step explanation:

Hope this helps~ :D

8 0

3 years ago

Read 2 more answers