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3 years ago

Four pencils to six markers

Mathematics

2 answers:

Basile [38]3 years ago

8 0

Answer:

4:6

2:3

2:3 would be the ratio

Step-by-step explanation:

I'm not sure if this is what you need but ya know hope it helps

charle [14.2K]3 years ago

6 0

The ratio is 4:6

I’m not really sure what the question was

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Hatshy [7]

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Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

$f_Y(y ) = \frac{1}{5-1} = \frac{1}{4}$ if $1\leq y \leq 5$ and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

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Also, we are given that $C = 50+3Y+9Y^2$ . Recall the following properties of the expected value. If X,Y are random variables, then

$E(a+bX+cY) = a+bE(X)+cE(Y)$

Then, using this property we have that $E[C] = 50+3E[Y]+ 9E[Y^2]$ .

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

$E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3$

$E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}$

Then

$E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152$

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3 years ago