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iren [92.7K]
3 years ago
5

How to you solve -15=y-62

Mathematics
1 answer:
sdas [7]3 years ago
7 0
First move the (Y) to the other side of the equal sign,
- y - 15 = - 62
Then subtract add (15) on both sides
- y = - 47
Since a variable can't be negative just divide by (-1) on both sides,
\frac{ - y}{ - 1} = \frac{ - 47}{ - 1}
The answer would be,
y = 47
Hope this helped
:D
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galben [10]

Answer:

30+30=11

Step-by-step explanation:

so bye

7 0
3 years ago
Translate into an inequality.
Alex787 [66]
5x-1 is greater than or equal to -11
5 0
3 years ago
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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
HELP ASAP
denis-greek [22]
A) I included a graph, look below.

B)

Input the y in x = y + 3.

x = (-4x - 3) + 3
x = -4x + 0
Add 4x to both sides.
5x = 0
Divide both sides by 5.
x = 0

Input that x value in y = -4x - 3

y = -4(0) - 3
y = 0 - 3
y = -3

(0, -3)

C)

Convert both equations to Standard Form.

x = y + 3
Subtract y from both sides.
x - y = 3

y = -4x - 3
Add 4x to both sides.
4x + y = -3

Add the equations together.

4x + y = -3
x - y = 3
equals
5x = 0
Divide both sides by 5.
x = 0

Input that into one of the original equations.

0 = y + 3
Subtract 3 from both sides.
-3 = y

(0, -3)

7 0
3 years ago
Ramon has started a workout program. On Monday, he did 10 push-ups. On Tuesday, he did 13 push-
nevsk [136]

Answer:

he will do 19 pushups

Step-by-step explanation:

He will do 19 push ups because the pattern Is going up by 3 each days so if you add 3 to 16 you will get 19!!!!!

Hope This Helped :)

7 0
3 years ago
Read 2 more answers
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