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3 years ago

13

each class AT BRIARWOOD ELEMentary collected at least 54 cans during the food drive. If there were 29 classes in the school what

was the least number of cans collected???

2 answers:

sertanlavr [38]3 years ago

6 0

You multiply 54 by 29 which will leave you with 1566

Lorico [155]3 years ago

5 0

What you would need to do is multiply the two numbers.
54x29= 1566

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Please help me on this problem

Orlov [11]

Answer:

The Proof with Figure, Statement and Reasons is given below.

Step-by-step explanation:

Given:

∠DAF ≅ ∠EBF,

DF ≅ FE

Prove:

Δ ADF ≅ Δ BFE

Proof:

Statements Reasons

a. ∠DAF ≅ ∠EBF ...........................Given

<u>48. ∠DFA ≅ ∠EFB </u>..........................Vertical Angles are congruent

DF ≅ FE ..........................<u>.49. Given</u>

50.<u> Δ ADF ≅ Δ BFE </u>.......................<u>..By Angle-Angle-Side test</u>

6 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:

$\hat a = max(X_i)$

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

4 years ago

erastova [34]

Based on the above scenario, the sales director conducted an <u>experiment</u>.

Because a treatment <u>was </u>applied to the sales representatives.

The the best statistical study for this situation because the sales director is trying to establish <u>correlation</u>.

<h3>What is an experiment?</h3>

An experiment is known to be a method that is et up with the aim to test a given hypothesis as a component of a scientific method.

In an experiment, there are the two key variables which are the independent and dependent variables. The independent variable is known to be a controlled or it is tested against the dependent variable.

See full question

Part A.

The sales director conducted a.) an experiment b.) a survey c.)an observational study because a treatment a.) was not b.)was applied to the sales representatives.

This is the best statistical study for this situation because the sales director is trying to establish a.) causality b.) correlation.

Learn more about experiment from

brainly.com/question/17274244

5 0

2 years ago

I need help don’t know how to do the problem

Triss [41]

Answer:

1/9

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

HELP PLEASE ANYONE IVE BEEN ASKING THE SAME QUESTION FOR SO LONG OMG

Gala2k [10]

Answer:

c

Step-by-step explanation:

C says that w must be greater than or equal to 452.2 and less than or equal to 455.0

8 0

2 years ago