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son4ous [18]
3 years ago
13

PLEASE HELP IM GONNA CRY

Mathematics
1 answer:
djyliett [7]3 years ago
5 0
If you draw a vertical line through the curve and it has 2 or more intersection points, the curve does not represent a function. The right graph is not a function. The left one is.
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Help me answer the question pleaseee!!!
vfiekz [6]

Option A,B,C,D,F are False

Option E is True

Step-by-step explanation:

Finding graphs of the equation: y= -4x+6

Checking all the options.

A) (-4,-10)

Plugging value of x as -4 and y as -10

-10=(-4)(-4)+6

-10 = 16+6

-10 ≠ 22

So, both sides of the equation will not be equal.

B) (-4,-22)

Plugging value of x as -4 and y as -22

-22=(-4)(-4)+6

-22=16+6

-22 ≠ 22

So, both sides of the equation will not be equal.

C) (1,10)

Plugging value of x as 1 and y as 10

10=(-4)(1)+6

10 = -4+6

10 ≠ 2

So, both sides of the equation will not be equal.

D) (1,-2)

Plugging value of x as 1 and y as -2

-2=(-4)(1)+6

-2 = -4+6

-2 ≠ 2

So, both sides of the equation will not be equal.

E) (4,-10)

Plugging value of x as 4 and y as -10

-10=(-4)(4)+6

-10 = -16+6

-10 = -10

So, both sides of the equation will be equal.

F) (4,-22)

Plugging value of x as 4 and y as -22

-22=(-4)(4)+6

-22 = -16+6

-22 ≠ -10

So, both sides of the equation will not be equal.

So, Option A,B,C,D,F are False

Option E is True

Keywords: Graph of equations

Learn more about Graph of equations at:

  • brainly.com/question/10879401
  • brainly.com/question/7490805
  • brainly.com/question/11150876

#learnwithBrainly

3 0
3 years ago
Choose the expression that best represents the product of the width and the height
nexus9112 [7]
The length of a rectangle is 12 inches more than its width. What is the width of the rectangle if the perimeter is 42 inches?

x best represents .

x + 12 best represents
6 0
2 years ago
Where would fraction<br> 22/11 belong on a #6 number line
pickupchik [31]
22/11 is equal to 2
So, it would be where 2 is on the number line.
6 0
3 years ago
A=340
ASHA 777 [7]

Answer:

  16666.25

Step-by-step explanation:

For rounded values, the minimum value they could have been rounded from is the given value less half its least-significant digit.

That means the minimums are ...

  a: 340 ⇒ 340 -5 = 335

  b: 49.8 ⇒ 49.8 -0.05 = 49.75

Then the minimum product would be ...

  (335) × (49.75) = 16666.25

8 0
1 year ago
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
3 years ago
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