All categories

3 years ago

Are -3.5 and -14/4 the same value? plzzz i need helpppp

Mathematics

2 answers:

mafiozo [28]3 years ago

4 0

Answer: Yes, they both equal -7/2

Step-by-step explanation:

AleksAgata [21]3 years ago

4 0

Answer:

Yes it is

Step-by-step explanation:

when you dived -14 by 4 you will get -3.5

You might be interested in

Identify the terms of the expression 9a+4b-18

madreJ [45]

Hey there!

We may ask. What exactly is a term anyways.

Well, a term is a single number, a variable, or a number and a variable that would have to be multiply together. This is what a term would be.

Therefore, as we look above, we see the following equation.

9a+4b-18

18 is a term

4b is a term

9a is a term

This whole thing, would be called an expression, as this is expression something. Perhaps the amount of students who ate ice-cream at a football game, or something similar.

I hope you found this helpful! :)

6 0

3 years ago

6 1/2 how many inches

sveticcg [70]

Answer:

78 inches

Step-by-step explanation:

12 TIMES 6 = 72

72+6=78

6 0

3 years ago

Solve 8x2 + 6x + 5 = 0

MariettaO [177]

8 x 2 + 6x + 5=0
16 + 6x + 5=0
6x+21=0
6x+21-21=0-21
6x=-21
6x/6=-21/6
x=-7/2

5 0

3 years ago

Plzz answer if u know it

liraira [26]

Answer:

A . 60

Step-by-step explanation:

30 x 2 = 60

6 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