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2 years ago

El producto de dos números pares

Mathematics

1 answer:

aliina [53]2 years ago

5 0

Answer:

Del Planeta ecuacion

Step-by-step explanation:

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Under his cell phone plan, Cameron pays a flat cost of $41 per month and $5 per gigabyte. He wants to keep his bill under $65 pe

hammer [34]

Answer- 512 (hope this helps)

6 0

3 years ago

Ten points QRSTUVWXY and Z are equally and consecutively spaced on a circle.

laiz [17]

The size of ∠QTW in degrees is 72°. See the explanation below. This is based on the concept of opposite sides in a quadrilateral.

<h3>What is the logic leading to the above result?</h3>

QSUWY which is a pentagon is regular that mean:

∠QYW = 108°

The quadrilateral QTWY is a cyclic one. This means that we can apply the rule that states that opposites sides must add up to 180°

Thus, ∠QTW =

180 - 108

= 72°

<h3>What is a quadrilateral?</h3>

A closed quadrilateral has four sides, four vertices, and four angles. It is a form of polygon.

In order to create it, four non-collinear points are joined. Quadrilaterals always have a total internal angle of 360 degrees.

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Full Question:

See the attached image for missing details.

4 0

2 years ago

Suppose that a box contains 8 cameras and that 4 of them are defective. A sample of 2 cameras is selected at random with replace

Dafna1 [17]

The Expected value of XX is 1.00.

Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.

The probability distribution of the hypergeometric is as follows:

$P(x,N,n,M)=\frac{\left(\begin{array}{l}M\\ x\end{array}\right)\left(\begin{array}{l}N-M\\ n-x\end{array}\right)}{\left(\begin{array}{l} N\\ n\end{array}\right)}$

Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

The probability distribution for X is obtained as below:

From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.

Here, M = 4, n=2 and N=8

The probability distribution of X is obtained below:

The probability distribution of X is,

$P(X=x)=\frac{\left(\begin{array}{l}5\\ x\end{array}\right)\left(\begin{array}{l}8-5\\ 2-x\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}$

The probability distribution of X when X=0 is

$\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

The probability distribution of X when X=1 is

$\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end$

The probability distribution of X when X=2 is

$\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

Use E(X)=∑xP(x) to find the expected values of a random variable X.

The expected values of a random variable X is obtained as shown below:

The expected value of X is,

E(X)=∑xP(x-X)

E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]

E(X)=[0+0.57+0.42]

E(X)=0.99≈1

Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.

Learn about Binomial probability distribution from here brainly.com/question/10559687

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8 0

2 years ago

I keep getting this wrong and I only have one chance to get it right. Please help! It's timed aswell.

allsm [11]

PIC 1 - floor function

PIC 2 - 1,1

PIC 3 - -1,1

PIC 4 - f(x)=[1/2]

PIC 5 - f(x)=1/5

7 0

3 years ago

Please help me and explain!

Sindrei [870]

Answer:

Step-by-step explanation:

See the attached document.

6 0

3 years ago