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Airida [17]
3 years ago
12

“I think I can make it. I’ll just jump.” says Denis, flashing a grin as he looks to the neighboring rooftop.

Mathematics
1 answer:
Elis [28]3 years ago
3 0

Answer: 5.3 m

Step-by-step explanation:

(i don’t know because i got it wrong but after I got it wrong it told me that answer was correct just trying to help you out lol)

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Dazuri saw the following advertisement at a sports store how much does each soccer ball cost
Nadusha1986 [10]

Based on the advertisement seen by Dazuri, the cost of each soccer ball is $22.50

<h3>How to find the cost?</h3>

The advertisement says that the original cost of the ball is $30 but there is a 25% discount.

The price of each soccer ball is therefore reduced by the 25% discount on the ball.

The amount that each soccer ball costs is therefore:
= 30 - ( 30 x 25%)

= 30 - 7.5

= $22.50

Rest of the question is:
Soccer balls on sale at a 25% discount. Prices go back to $30 in a week. Don't miss out.

Find out more on cost of soccer ball at brainly.com/question/2227883

#SPJ1

4 0
1 year ago
There are 16 students in a class with wooden pencils. There are 20 students in a class with mechanical pencils. If a school has
svet-max [94.6K]

Answer:

maybe 100 cheak it please

5 0
3 years ago
Read 2 more answers
A sphere has a volume of 1,436 in3. What is its radius?
Gwar [14]

Answer:

do the reverse of pie r squared

Step-by-step explanation:

4 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
Hi! I need help on #12 A, and B! Just kinda confused so if u could explain in a simple way that’d be great!
Vinvika [58]

Assume that the bank balance started at zero. Enter the income and outgo in chronological order: $300 - $50 - $75 + $225

Find the sum. This sum will represent the amount left in the account after the $225 deposit:

$250 - $75 + $225, or

$175 + $225 = $400 The final balance will be $400.

5 0
4 years ago
Read 2 more answers
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