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FinnZ [79.3K]
3 years ago
13

Identify square root of 2 as either rational or irrational and approximate to the tenths place. Rational: square root of 2 ≈ 1.5

Irrational: square root of 2 ≈ 1.5 Rational: square root of 2 ≈ 1.4 Irrational: square root of 2 ≈ 1.4
PLEASE ANSWER FAST will give gris Liset
Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0
The square root of 2 is irrational since it is not a perfect square. also <span>Irrational: square root of 2 ≈ 1.4</span>
faltersainse [42]3 years ago
7 0

Answer:

Last option.

Step-by-step explanation:

\sqrt{2} is an irrational number because has infinite decimal values. So it can't be represented by a fraction.

\sqrt{2}= 1.4142... = 1.4.

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The probability of being a universal donor is 6% (O-negative-blood type). Suppose that 6 people come to a blood drive.
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Answer:

babala

Step-by-step explanation:

The probability of being a universal donor is 6% (O-negative-blood type). Suppose that 6 people come to a blood drive.

a) What are the mean and standard deviation of the number of universal donors among the 6 people?

b) What is the probability that there are exactly three universal donors?

7 0
3 years ago
Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari
Ulleksa [173]

Answer:

(a)E[X+Y]=E[X]+E[Y]

(b)Var(X+Y)=Var(X)+Var(Y)

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).

Since f(X,Y)=X+Y

E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).

Let us look at the first of these sums.

\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].

Similarly,

\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].

Combining these two gives the formula:

\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)

Therefore:

E[X+Y]=E[X]+E[Y] \text{  as required.}

(b)We  want to show that if X and Y are independent random variables, then:

Var(X+Y)=Var(X)+Var(Y)

By definition of Variance, we have that:

Var(X+Y)=E(X+Y-E[X+Y]^2)

=E[(X-\mu_X  +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2  +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2  +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2  +E[Y- E(Y)]^2+2Cov (X,Y)

Since X and Y are independent, Cov(X,Y)=0

=Var(X)+Var(Y)

Therefore as required:

Var(X+Y)=Var(X)+Var(Y)

7 0
3 years ago
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