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2 years ago

Please hurry I only have 5 minutes left!! I will mark brainliest answer!!

Mathematics

2 answers:

Illusion [34]2 years ago

7 0

Answer:

Step-by-step explanation:

8_murik_8 [283]2 years ago

4 0

Answer:

It is B :3

Step-by-step explanation:

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Please answer fast its timed

Alborosie

Answer:

f(5) = 22

f(9) = 34

Step-by-step explanation:

Plug in 5 and 9 separately for the equation of the x values. Multiply them each and add 7 to get these values 22 and 34.

4 0

3 years ago

The Wonderful Widget Company's earnings have declined to the point where they need to reduce the number of employees from 1500 1

AleksandrR [38]

Answer:6%

Step-by-step explanation:

Decrease = 1500 - 1410= 90

% decrease = (90/1500)*100%

= 6%

7 0

2 years ago

SOMEONE HELP ME this is due in like less than 15 mins

zmey [24]

Answer:

for the first one I think it is -5/-2 second is 1/-4 third one is 2/1 this may not be correct

Step-by-step explanation:

5 0

3 years ago

Slove:33/29=46/x. Try to find the answer.

Tju [1.3M]

Answer:

$x=40\frac{14}{33}$

Step-by-step explanation:

we have

$\frac{33}{29}=\frac{46}{x}$

Solve for x

Multiply in cross

$(33)(x)=(46)(29)$

Divide by 33 both sides (Division Property of Equality)

$x=46(29)/33$

$x=\frac{1,334}{33}$

Convert to mixed number

$x=\frac{1,334}{33}=\frac{1,320}{33}+\frac{14}{33}=40\frac{14}{33}$

5 0

2 years ago

Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g

Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

$E[g(x)] = \sum g(x)p(x)$

So, $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

$g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also, $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

$g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0

2 years ago