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denis-greek [22]
3 years ago
7

there are 81 students in green t-shirts and 54 in white t-shirts and 27 in orange t-shirts. the teacher is planning to arrange t

hem in such a way that each row has same colour t-shirts and all rows have equal number of students. find the greatest number of students
Mathematics
2 answers:
Eduardwww [97]3 years ago
6 0

Answer:

54

Step-by-step explanation:

because it was a logic question

krok68 [10]3 years ago
3 0

Answer:

ancient Greek literature

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I need an answer for the empty spaces
abruzzese [7]

Answer:

4, +, 6

Step-by-step explanation:

3 0
3 years ago
There are 32 girls and 17 boys what percent of the class are girls
gulaghasi [49]
Given that there are 32 girls and 17 boys, the total number of students in the class is 49. In order for us to know the percent of the girls in class, what we are going to do is to simply divide 32 by 49 and then multiply by 100 and the answer is 65.3%. Hope this answers your question.
8 0
3 years ago
The graph of a polynomial S(x) = (2x - 3)(x – 4)(x+3) has x-intercepts at 3.x values.
dezoksy [38]

Answer:

Step-by-step explanation:

x-intercepts exist where y is equal to 0. Where y is equal to 0 is where the graph goes through the x-axis. Our x-intercepts are (2x-3), (x + 3), and (x-4). Again, since x-intercepts exist where y = 0, then by the Zero Product Property, 2x - 3 = 0, x - 4 = 0, and x + 3 = 0. In the first x-intercept:

2x - 3 = 0 and

2x = 3 so

x = 3/2

In the second:

x - 4 = 0 so

x = 4

In the third:

x + 3 = 0 so

x = -3

So the x-intercepts in the correct order are x = 3/2, 4, -3

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
3 years ago
Type the simple subject of this sentence.the plane leaves at 6:16 and arrives at 10:50.
Jet001 [13]
Shouldn't this be in the language section? But okay. I would say the simple subject would be "plane"

5 0
3 years ago
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