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3 years ago

11

HURRY ASAP NO JOKE 2 MINUTES PLZ HURRY ASAP PLZZZZZ HELP

1 answer:

loris [4]3 years ago

5 0

$2 \frac{3}{4} * \frac{4}{5}$ is the same as $\frac{11}{4} * \frac{4}{5}$ . The 4s will cancel leaving you with $\frac{11}{5}$ or $2 \frac{1}{5}$ , your 3rd option.

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3 years ago

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CAB is similar to EFD. What is the length of DE?

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3 years ago

Can bbn I please have help

Sveta_85 [38]

Answer:

question 9

= –12

while question 11=, x= –72

Step-by-step explanation:

question 9=

u= –16+4= –12

question 11=

cross multiplication

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3 0

3 years ago

4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean

Ede4ka [16]

Adding 2 to each value of the random variable $X$ makes a new random variable $X+2$ . Its mean would be

$E[X+2]=E[X]+E[2]=E[X]+2$

since expectation is linear, and the expected value of a constant is that constant. $E[X]$ is the mean of $X$ , so the new mean would be

$E[X+2]=10+2=12$

The variance of a random variable $X$ is

$V[X]=E[X^2]-E[X]^2$

so the variance of $X+2$ would be

$V[X+2]=E[(X+2)^2]-E[X+2]^2$

We already know $E[X+2]=12$ , so simplifying above, we get

$V[X+2]=E[X^2+4X+4]-12^2$

$V[X+2]=E[X^2]+4E[X]+4-12^2$

$V[X+2]=(V[X]+E[X]^2)+4E[X]-140$

Standard deviation is the square root of variance, so $V[X]=3^2=9$ .

$\implies V[X+2]=(9+10^2)+4(10)-140=9$

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of $aX+b$ for $a,b\in\mathbb R$ is

$V[aX+b]=a^2V[X]+b^2V[1]$

but the variance of a constant is 0. In this case, $a=1$ , so we're left with $V[X+2]=V[X]$ , as expected.

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3 years ago

F(x)=2x-6 find f(2) please help

vladimir1956 [14]

Answer:

Step-by-step explanation:

Putting value of x = 2 in the equation

F(2) = 2(2) - 6

= 4 - 6

= - 2

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2 years ago