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2 years ago

Evaluate the expression. StartFraction 9 factorial Over 3 factorial EndFraction

Mathematics

2 answers:

Kaylis [27]2 years ago

7 0

Answer:

The answer is C

Step-by-step explanation:

I just did it and got it right.

Verizon [17]2 years ago

5 0

Answer:

The answer you should get is 9:3

Step-by-step explanation:

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2. A lunch stand makes s.75 profit on each chef's salad and $1.20 profit on each Caesar salad. On a typical weekday, it sells be

kow [346]

Let's let

x = the number of chef salads, x>=0

y = the number of Caesar salads, y>=0

The constrains are:

40 <= x <= 60

35 <= y <= 50

x + y <= 100

The objective function here is F(x, y) = 0.75x + 1.20y

The corner points are (40, 35), (60, 35), (60, 40), (50, 50) and (40, 50).

F (40, 35) = 0.75*40 + 1.20*35 = $72

F (60, 35) = 0.70*60 + 1.20*35 = $84

F (60, 40) = 0.75*60 + 1.20*40 = $93

F (50, 50) = 0.75*50 + 1.20*50 = $97.50

F (40, 50) = 0.75*40 + 1.20*50 = $90

Thus, we conclude to maximize the profit 50 Chef and 50 Caesar salads should be prepared.

8 0

3 years ago

Which of the following is an expression? *

Yanka [14]

Answer:

Step-by-step explanation:

An expression does not have an equals sign

An equation has an equals sign

4x = 12 equation

25 - 12 = 13 equation

7x - 3 + 2y =29 equation

3m expression

4 0

2 years ago

HELP PLEASE Find the equation of the line The line passes through the points (2, 3) and (-3, -2)

Vladimir79 [104]

Y= x+1

y-3 = (-2-3)/(-3-2) (x-2)
y-3 = 1 (x-2)
y = x-2+3
y=x+1

Mark brainliest please

7 0

2 years ago

Which pair of numbers has a GCF of 3? Group of answer choices (A) 3 and 18 (B) 8 and 24 (C)12 and 18 (D) 1 and 3

Arte-miy333 [17]

Answer:

A, B and C

Step-by-step explanation:

A) 3 and 18 = has 3 as factor

B) 8 and 24 = has 3 as factor

C) 12 and 18 = has 3 as factor

D) 1 and 3 = has as factor of 1 and 3

so the answer is A, B and C

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

2 years ago