Look at the picture and you can get the brainliest if you get it right no links

Select the correct answer from the drop-down menu.

Contact [7]

Answer:

The correct option is C.

C) -x + 2

Step-by-step explanation:

In mathematics, the term which is being divided is called a dividend. The term by which the dividend is getting divided is called a divisor. The quotient is the term we get after dividing the dividend by a divisor.

Here in this case, the dividend is written.

-x² x x

which in the equation form is written as

-x² + x + x

add the x terms to simplify

-x² + 2x

The divisor is given as x, divide the equation by x:

(-x² + 2x)/x

(-x²/x) + (2x/x)

-x+2

Which is the quotient.

4 0

3 years ago

Donatello Co. has identified an activity cost pool to which it has allocated estimated overhead of $9,600,000. It has determined

ahrayia [7]

Answer:

option (a) Widgets $2,400,000, Gadgets $1,800,000, Targets $5,400,000

Step-by-step explanation:

Data provided in the question:

Allocated estimated overhead = $9,600,000

Expected use of cost drivers for the activity = 800,000 inspections.

Inspections required by Widgets = 200,000

Inspections required by Gadgets = 150,000

Inspections required by Targets = 450,000

Therefore,

Total overhead per activity = $\frac{\textup{estimated overhead}}{\textup{Total activity}}$

= $\frac{\textup{9,600,000}}{\textup{800,000 }}$

= $12 per activity

Thus,

Overhead allocated to Widgets = 200,000 × 12 = $2,400,000

Overhead allocated to Gadgets = 150,000 × 12 = $1,800,000

Overhead allocated to Target = 450,000 × 12 = $5,400,000

Hence, The correct answer is option (a)

8 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

Help plzz (30) points for both of the question-plz help - this is life or dead answer

Elodia [21]

Answer:

The answer is B) 11.

Step-by-step explanation:

We can solve this by making an expression to represent the situation. Let x = the number of student books in the box (what we're looking for):

$2.3x + 4.3 + 0.4 = 30$

$2.3x + 4.7 = 30$

$2.3x + 4.7 - 4.7 = 30 - 4.7$

$2.3x = 25.3$

$2.3x/2.3 = 25.3/2.3$

$x = 11$

8 0

3 years ago

A contractor is calculating the number of decorative stones for a new rectangular patio. The bids always include a count for the

babunello [35]

Answer:

The number of edge stones needed has a constant rate of change, but the number of stones for the center does not.

Step-by-step explanation:

The complete question is shown in the picture attached below.

Two functions E(x) and C(x) are given in the table along with some of the function values. We have to identify if any of these functions show constant rate of change or not.

By constant rate of change we mean that the slope is constant or in other words, a linear relationship is shown by the function. For a Linear function, the first differences of the function values are same.

By first differences we mean the difference between two consecutive output values. We can see that difference between consecutive input values is constant i.e. 1. If the difference between consecutive output values of any of the functions is same, then that function will be a Linear Function and, therefore, the rate of change for that function will be constant.

Lets analyze E(x) first. The output values are:

34, 52, 70 and 88

The difference between consecutive output values is:

18, 18 and 18

Since, this difference is constant, we can conclude that E(x) is a Linear function with a constant rate of change.

Now lets analyze C(x). The output values are:

62, 133, 260 and 435

The difference between consecutive output values is:

71, 127 and 175

Since, the differences are not constant, C(x) is not a Linear Function and , therefore, the rate of change of C(x) is not constant.

Conclusion:

The number of edge stones which is represented by E(x) has a constant rate of change, but the number of stones for center which is represented by C(x) does not have constant rate of change. This makes the first option our correct answer.

3 0

2 years ago