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mezya [45]
3 years ago
9

paul earned 175 for working 5 days last week. He earned the same amount each day. What was pauls pay rate?

Mathematics
2 answers:
andre [41]3 years ago
8 0

Answer:

His pay rate is $35.

Step-by-step explanation:

blondinia [14]3 years ago
7 0

Answer: His pay rate is $35

Step-by-step explanation:

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2x + 9 = 23 solve by trial
Allisa [31]

Answer:

i tried it by trial and error, and my work is below.

Step-by-step explanation:

2x + 9 = 23

subtract 9 from both sides

2x = 14

guess.... lol

x=7

8 0
3 years ago
Pls help me with this
Marina CMI [18]

Answer: Can I see the whole questions?

Step-by-step explanation:

6 0
2 years ago
Myles said that 5 x 0.13 is 6.5. Do you agree? Explain.<br><br> Brainliest if correct!
77julia77 [94]
No he is not, 10*0.13=1.3, so 5*0.13=0.65
6 0
3 years ago
Read 2 more answers
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
What is the simplified form of the following expression?<br>  
ra1l [238]
(4/7)^3=.1865

6xyz/2xz simplifies to 3y
6 0
3 years ago
Read 2 more answers
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