All categories

3 years ago

While solving an equation, if the variable term becomes zero, and the equation makes a true statement, then the solution is

Mathematics

1 answer:

sergiy2304 [10]3 years ago

4 0

Answer:

the y-intercept

Step-by-step explanation:

For example, the equation is 2x+3=3

2x+3=3

2x=0

x=0

You might be interested in

Find the value of the missing segment below please.

kogti [31]

Answer:

Um it might be 1 but feel free to correct me if I'm wrong

7 0

3 years 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:

$\hat a = max(X_i)$

$E(a) - a= 0$ 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

4 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS PLS ASAP

Alenkasestr [34]

Answer:

(0, 6)

explanation:

original coordinates of C: (5, -6)

if reflected over x-axis, then use the formula: (x,y) → (x, -y)

new coordinates: (5, 6)

then if translated -5 horizontally, there will be change in x axis

new coordinates: (5-5, 6) → (0, 6)

3 0

3 years ago

Kindly answer this one. Thank you.

steposvetlana [31]

Answer:

See below.

Step-by-step explanation:

I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.

♦ The midline's length is the average of the lengths of the top and bottom parallel sides.

$AL=\frac{OR+CE}{2}$

Use this equation and substitute values given in each problem, then solve for the missing information.

1. AL = x, CE = 9, OR = 5

$x=\frac{9+5}{2}=7$

2. AL = <em>m</em> - 4, CE = 15, OR = 17

$m-4=\frac{15+17}{2}=16\\\\m-4=16\\\\\\m=20\\\\AL=20-4=16$

3. OR = y + 5, AL = 15, CE = 18

$15=\frac{(y+5)+18}{2}\\\\15=\frac{y+23}{2}\\\\30=y+23\\\\7=y\\\\OR = 7+5=12$

4 0

3 years ago

What is the standard form of the equation of the circle in the graph?

jonny [76]

What graph......................

3 0

3 years ago