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3 years ago

Is the relation below a function? -3 -1 -1 3 у 0 1 2 3

Mathematics

2 answers:

wolverine [178]3 years ago

8 0

Answer:

<h3>I would say no!</h3>

Step-by-step explanation:

<h3>cause we cannot determine the rule of mapping so y is no an image of x</h3>

qaws [65]3 years ago

5 0

Answer:

I believe it would be "y"

Step-by-step explanation:

Function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

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Find the LCD of the fractions, and then simplify the expression. Assume that no denominator equals zero. 5/(3x^2y) - 4/(6xy^3

tamaranim1 [39]

The LCD = 6x^2y^3 ( because LCD of 3 and 6 = 6, LCD of x^2 and x = x^2 and LCD of y and y^3 = y^3)

now divide 3x^2y into the LCD then multiply this by 5 to get the first term in the numerator and do similar process to get second term, so we get:-

5(2y^2) - 4(x)
------------------
6x^2y^3

= 2( 5y^2 - 2x)
-----------------
6x^2y^3

= 5y^2 - 2x
-----------
3x^2y^3

5 0

3 years ago

If f(x) = V7x + 31 , find f(0).

Arte-miy333 [17]

Answer:

f(0) = 31

Step-by-step explanation:

We are given the following function:

$f(x) = \sqrt{7x} + 31$

f(0).

This is the value of f when $x = 0$ . So

$f(0) = \sqrt{7*0} + 31 = \sqrt{0} + 31 = 0 + 31 = 31$

So f(0) = 31

3 0

3 years ago

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

Plss help me<br><br> NO LINKS

Degger [83]

Answer:

17 square yards

Step-by-step explanation:

split it so you have a big rectangle and a small one. find the areas of both

the big one was 5(3)= 15, the second was 1(2)=2. add those together.

5 0

3 years ago

Read 2 more answers

3x+18x+5x-2=180<br>what is the angle measurement for:<br>3x,18x,and 5x,-2?<br>

lozanna [386]

Answer:

3x = -18 (x = -6)

18x = 108 (x = 11)

5x = 90 (x = 18)

Step-by-step explanation:

7 0

3 years ago