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3 years ago

Which is the solution for log(3x - 1) + log 2 = log 4 + log (x + 2) ?

Mathematics

2 answers:

Lemur [1.5K]3 years ago

6 0

Answer:x=1

Step-by-step explanation:

log(3x-1)+log2=log4+log(x+2)

log(2(3x-1))=log(4(x+2))

Cancel log on both sides

2(3x-1)=4(x+2)

Open brackets

6x-2=4x+8

Collect like terms

6x-4x=8+2

10x=10

Divide both sides by 10

10x/10=10/10

x=1

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

x=5

Step-by-step explanation:

log(3x-1)+log2=log4+log(x+2)

log(2(3x-1))=log(4(x+2))

Cancel log on both sides

2(3x-1)=4(x+2)

Open brackets

6x-2=4x+8

Collect like terms

6x-4x=8+2

2x=10

Divide both sides by 2

2x/2=10/2

x=5

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mihalych1998 [28]

Answer:

John and Pam are paid $8.5 for each hour worked. John's share of the money is $29.75.

Step-by-step explanation:

Let x = the hourly salary. John worked for 3.5 hrs, and Pam for two. We can represent this using the equation:

3.5x + 2x = 46.75, where the coefficients equals the amount of hours.

Let's solve for x!

5.5x=46.75

x = 46.75/5.5 = 17/2 = 8.5

John and Pam are paid 8.5 dollars per each hour worked.

To figure out John's share of the money, we will multiply the wage by the hours he worked.

8.5 x 3.5 = 29.75

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printArray(arr[], integers)
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3 years ago

The sum of 14 and a number is equal to 17

alexdok [17]

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3 years ago

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djverab [1.8K]

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Step-by-step explanation:

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3 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

Which is the solution for log(3x - 1) + log 2 = log 4 + log (x + 2) ?​

Which is the solution for log(3x - 1) + log 2 = log 4 + log (x + 2) ?