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hoa [83]
3 years ago
13

I need help with both of the Questions.

Mathematics
2 answers:
Sphinxa [80]3 years ago
5 0

15) Basically, this year Chelsea is making 64 cups of soup, and since there are 16 cups in a gallon, Chelsea will make 4 gallons.

16) Option D

Katarina [22]3 years ago
5 0

15)

The answer is C.

16)

The answer is D.

Hope this helps!

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Laquita practiced 60 math facts. Kinman practiced 15 more facts than Laquita. How many math facts did they practice in all?
Afina-wow [57]

Answer:

They practiced a total of 75 math facts in all!

60 + 15 = 75 :D

5 0
2 years ago
Sunshine Rental Car charges a one-time fee of $30.15 per mile the yellow car agency charges a one-time fee of $20.25 per mile th
11111nata11111 [884]

Answer:

105 miles

Step-by-step explanation:

I used a calculator because it was multi choice until they matched up

5 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
Evaluate the expression 4ab, for a =2 and b = 5
yan [13]

Answer:

40

Step-by-step explanation:

4x2=8

8x5=40

4 0
2 years ago
Read 2 more answers
The question is in the picture
faust18 [17]

Answer:

the hypotenuse is 10

Step-by-step explanation:

6^2 + 8^2 = 100

square root 100 and its 10

7 0
3 years ago
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