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Darina [25.2K]
3 years ago
6

Pls help answer I’m thinking it’s negative but I’m not sure.

Mathematics
2 answers:
Serhud [2]3 years ago
5 0
The answer is negative
Digiron [165]3 years ago
3 0
Negative i believe is the correct answer
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Solvex2<br> - 4x=21 <br> using the zero product property.
Ierofanga [76]

Answer:

The answer is (x + 3) (x - 7) = 0

Step-by-step explanation:

x² - 4x = 21

x² - 4x - 21 = 0

x² - 7x + 3x - 21 = 0

x(x - 7) + 3(x - 7) = 0

(x + 3) (x - 7) = 0

Thus, The answer is (x + 3) (x - 7) = 0

<em><u>-TheUnknownScientist</u></em>

5 0
3 years ago
Can someone help me with this? Does it mean cup or is it a typo? Also I need help with the answer.
algol [13]

Answer:

Yes it really does mean cup.

Step-by-step explanation:

This can only contian 1 jar

12.5 cups are equal to 100 ounces

if there are 100 oz in that little jar then only 1 large jar can be filled by a lttile jar.

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3 years ago
Read 2 more answers
I need help with all of them.. Please help me
jarptica [38.1K]

Answer:

1. 8

2. 0.49

3. 7

4. 0.52

5. 0.43

6. 28

Step-by-step explanation:

I did the math you are welcome.

5 0
2 years ago
A painter can paint 350 square feet in 1.25 hours. What is the painting rate per square feet per hour?
Andrei [34K]

Answer:

280

Step-by-step explanation:

350÷5 will give you quarters. 350÷50=70. 70x4=280

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
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