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2 years ago

Solve for all possible values of x.√x-8-3=1Ox= -4O x = 8O x = 12Ox= 24

Mathematics

1 answer:

faust18 [17]2 years ago

5 0

Answer:

x = 24

Step-by-step explanation:

$\sqrt{x - 8} - 3 = 1 \\ \sqrt{x - 8} = 1 + 3 \\ \sqrt{x - 8} = 4 \\ squaring \: both \: sides \: \\ {( \sqrt{x - 8)} }^{2} = {4}^{2} \\ x - 8 = 16 \\ x = 16 + 8 \\ \huge \red{ \boxed{x = 24}}$

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Step-by-step explanation:

perimeter = side 1+side 2

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Regina, Phil, and Joseph each wrote expressions to represent their hourly earnings for an after-school job where h represents th

Vladimir79 [104]

Answer:

Disagree

On weeks when Joseph works extra 2/3 hours more than Phil, Phil and Joseph can earn the same amount of money

Step-by-step explanation:

The given expression for the hourly earnings of Regina, Phil and Joseph are as follows;

Regina: 6.50·h + 16

Phil: 3·(2.5·H + 5)

Joseph: 7.50·h + 10

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The hourly earnings for Phil equated to the hourly earnings of Joseph gives;

3·(2.5·H + 5) = 7.50·h + 10

Simplifying we get;

3 × 2.5·h + 3 × 5 = 7.50·h + 10

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Which gives

15 = 10 + 5 = 10 which is not correct.

Therefore, when both Phil and Joseph works the same number of hours, in the week, Phil will earn 5 units more Joseph

However, we have that, where Joseph works extra 5/7.5 or 2/3 hours more than Phil, every week, they will earn the same amount of money every week.

Therefore, it is possible for Phil and Joseph to earn the same amount of money when Joseph works extra 2/3 hours more than Phil which gives;

On weeks when Joseph works extra 2/3 hours more than Phil, Phil and Joseph can earn the same amount of money.

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3 years ago

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Step-by-step explanation:

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2 years ago

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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

Solve for all possible values of x.√x-8-3=1Ox= -4O x = 8O x = 12Ox= 24​

Solve for all possible values of x.√x-8-3=1Ox= -4O x = 8O x = 12Ox= 24