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Bingel [31]
2 years ago
9

True or False: The relation shown is a function. 20 POINTS!!

Mathematics
1 answer:
elixir [45]2 years ago
4 0

Answer:

True

Step-by-step explanation:

For a relation to be a function each value of x must pair up with one and only one value of

The relation shown is a function.

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About how many pounds are in 14 kilograms?
sesenic [268]

Answer:

308.647

Step-by-step explanation:

Its searchable pls mark brainliest.

6 0
2 years ago
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1/(x+4)(x-2)(x-5) excluded values
UNO [17]

Answer:

see explanation

Step-by-step explanation:

The excluded values are any values of x that make the function undefined

Given

\frac{1}{(x+4)(x-2)(x-5)}

The denominator of the rational function cannot be zero as this would make it undefined. Equating the denominator to zero and solving gives the values that x cannot be.

(x + 4)(x - 2)(x - 5) = 0

Equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

x - 2 = 0 ⇒ x = 2

x - 5 = 0 ⇒ x = 5

x = - 4, x = 2, x = 5 are the excluded values

4 0
2 years ago
Can someone help me with and and can you please write how you got it so I can understand
EastWind [94]

Answer:

12x + 4

Step-by-step explanation:

ok so a rectangle has 4 sides, 2 of which are called lengths, and 2 of which are widths.

length = 2x-3

width = 4x+5

the perimenter is the total of the 4 sides.

2(2x-3) = 4x-6 (the total for lengths)

2(4x+5) = 8x+10 (the total for widths)

add the 2 together:

12x+4

8 0
2 years ago
Read 2 more answers
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
The temperature in an oven changes from 350 degrees to 362 degrees. What is the percent increase in temperature, to the nearest
Aleksandr-060686 [28]

Answer:

it's 3.4%

Step-by-step explanation:

You take the difference of 362 and 350 (which is 12) and divide it by the starting number (350) getting you 3.4

5 0
2 years ago
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