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3 years ago

Log(4)+log(2)-log(5)

2 answers:

5 0

   log(4) + log(2) - log(5)

   = log(2²) + log(2) - log(5)

   = 2 log(2) + log(2) - log(5)

   = 3 log(2) - log(5)

   = log(2³) - log(5)

   = log (2³/5)

   = log (8/5)

   = log (1.6) = 0.2041... (rounded)

Naily [24]3 years ago

5 0

=log (x4) + =log (x2) - =log (x5)

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Kay scored above 90% on five of her math quizzes, 80% to 90% on fourteen of them, and 70% to 80% on two of them. What is the exp

Leya [2.2K]

In total, Kay took 5+14+2=21exams. she got 80%-90% on 14 of them. the probably is 14/21 which is 2/3. Therefore, the probability of Kay getting 80%-90% on her next exam is 2/3 or 66.6%

3 0

3 years ago

What is 27.773 to its nearest tenth

Olenka [21]

Check the hundredth position after the decimal point.

This = 7 so we add 1 to the tenth position so 7 becomes 7+1 = 8

Answer is 27.8 to nearest tenth

7 0

3 years ago

How tall is the flagpole?

Fudgin [204]

51 ft is the answer

Explanation

The hight of the man divided by the hight of school ( which is unknown )

Equal

The length of the shadow of the man divided by the shadow of the school

Cross multiplication and find the value of unknown x which is 51... and ya

4 0

3 years ago

If line GH is perpendicular to line HJ, then GJ is ____ HJ.

valentinak56 [21]

Option A: >

Solution:

Given a triangle GHJ.

The line GH is perpendicular to line HJ.

This means the triangle is a right angled triangle.

In ΔGHJ, GH is the base of the triangle and

HJ is a height of the triangle.

Then the third side must be the hypotenuse of the right triangle.

We know that by the Pythagoras theorem,

$(\text {Hypotenuse})^2=(\text{Base})^2+(\text{Height})^2$

$(\text {GJ})^2=(\text {GH})^2+(\text {HJ})^2$

This clearly shows that the hypotenuse is greater than the height.

⇒ GJ > HJ

Option A: > is the correct answer.

If line GH is perpendicular to line HJ, then GJ is > HJ.

3 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