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densk [106]
3 years ago
13

Find the GCF of -10c2d and 15cd2.

Mathematics
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

-5cd

Step-by-step explanation:

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morpeh [17]

Answer:

This might not help but I know 10 to the power of 3 is 1000 and the answer might be 10×10×10. 10×10×10 is equal to 1000 and is the same as 10 to the power of 3.

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A dime is 1/2 inch wide if you put 5 dimes end to end, how long will they be begginning to end
Contact [7]
If you put 5 which means its 2 1/2 ( I think) Because two 1/2 makes one.
8 0
3 years ago
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JUST 1 PLEASEEEEEEEEEE AND HOW DO YOU KNOW!!!!! PLEASEEEEEEEEEEEEE HELLLPPPPPPPPPPPP MEEEEEEEEEEEEEE
Lemur [1.5K]

Answer:

I can't see three but I'll answer the first two.

Step-by-step explanation:

Number one: The first, third, and fifth options are correct,

Number two: A is correct because if you divide 3/4 you get 0.75. Check by multiplying 0.75 for one by 4 and you get three dollars.

So for three, one pound of pears costs two dollars. That is consistent with the graphs. Pick the answer that states that the best.

5 0
2 years ago
Factories A and B produce computers. Factory A produces 3 times as many computers as factory B. The probability that an item pro
OleMash [197]

Answer:

P(A∣D) = 0.667

Step-by-step explanation:

We are given;

P(A) = 3P(B)

P(D|A) = 0.03

P(D|B) = 0.045

Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.

Using Bayes' Rule and Law of Total Probability, we will get;

P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]

Plugging in the relevant values, we have;

P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]

P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]

P(B) will cancel out to give;

P(A∣D) = 0.09/0.135

P(A∣D) = 0.667

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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