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3 years ago

What is the solution to -10 + p = -19

Mathematics

2 answers:

Law Incorporation [45]3 years ago

8 0

Answer:

p = -9 is the answer

9966 [12]3 years ago

3 0

Answer:

P = -9

Step-by-step explanation:

-10 + P = -19

(-10 + 10) + P = (-19 + 10)

P = -9

Hope I got it right, if so, hopefully it helped...

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Thank you so much for your help

lana [24]

Answer:

1.1x

Step-by-step explanation:

that is the procedure above

4 0

2 years ago

What is the factorization of the trinomial below? 3x3 + 6x2 - 24x

Allisa [31]

3x^3 + 6x^2 - 24x

= 3x(x^2 + 2x - 8)

= 3x(x + 4)(x - 2)

Hope it helps.

8 0

2 years ago

Read 2 more answers

A jar contains 30 red, 30 blue, and 60

kiruha [24]

The probability of picking red button is $\frac{1}{4}$

<u>Solution:</u>

Given, A jar contains 30 red, 30 blue, and 60 white buttons.

You pick one button at a random choice,

We have to find the probability that it is red button.

Now, total number of buttons = 30 + 30 + 60 = 120 buttons.

Number of favourable cases for red button = 30 red buttons.

The probability of an event is given as:

$\text { probability of a event }=\frac{\text { favourable ways to pick red }}{\text { total possiblities }}$

$=\frac{30}{120}=\frac{1}{4}$

Thus the probability that it is red is $\frac{1}{4}$

3 0

3 years ago

P-3 1/5=7 1/3 I need help

alekssr [168]

We have:

$p-3\frac{1}{5} =7\frac{1}{3} => p=7\frac{1}{3}+3\frac{1}{5}= 10\frac{8}{15}$

ok done. Thank to me :>

7 0

2 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