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Drupady [299]
3 years ago
13

Brandon is running errands for his mother.

Mathematics
1 answer:
Rudiy273 years ago
3 0

Answer:

He is correct

Step-by-step explanation:

You have to make both 1/4 and 2/3 have the same denominator to be able to add for the total distance walked. Since 12 is their LCM they will now have a denominator of 12.1/4 then turn into 3/12 and 2/3 turns into 8/12. When you add 8/12 and 3/12 you should get 11/12

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Bodmass<br><br>-3[2/6+8{-9/3(8-5*3)-6}]​
Vadim26 [7]

9514 1404 393

Answer:

  -361

Step-by-step explanation:

Your calculator can tell you the result. It is -361.

Start with the inner parentheses and work outward. Do multiplication and division in the order shown, left to right, before addition or subtraction.

  -3[2/6+8{-9/3(8-5*3)-6}]​

  = -3[2/6+8{-9/3(8-15)-6}]​

  = -3[2/6+8{-9/3(-7)-6}]​

  = -3[2/6+8{-3(-7)-6}]​

  = -3[2/6+8{21-6}]​

  = -3[2/6+8{15}]

  = -3[1/3+8{15}]

  = -3[1/3+120]

  = -3[361/3]

  = -361

8 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)

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3 years ago
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saveliy_v [14]
Option C is correct.

70000 \times 15 \times 5 \div 100 \\  \\
SI =P×R×T÷100
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The image below is a triangle drawn inside a circle with center O: A triangle is shown inscribed inside a circle. The leg of the
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I cant answer without a pic ture plz snip it out or screen shot thx :)
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Step-by-step explanation:

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