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1 year ago

Estimate the value of √15 to the nearest tenth.

Mathematics

2 answers:

Marianna [84]1 year ago

5 0

Answer: 3.9 Hope this helps, please consider making me Brainliest.

Step-by-step explanation:

We can use our knowledge of perfect square (roots). The perfect square roots are 1, 4, 9, 16, 25, etc. The closest one that sqrt15 is near to is 16. The sqrt of 16 is 4, so if 15 + 1 = 16, the correct answer is 3.9. Hope this helps, please consider making me Brainliest.

Nadya [2.5K]1 year ago

3 0

Answer:

3.9

Step-by-step explanation:

√15 = 3.9, rounded.

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Please help, only the answer is required

Romashka [77]

Answer:

Infinitely many solutions.

Step-by-step explanation:

5x-8y=-11

-10x+16y=22

---------------------

2(5x-8y)=2(-11)

-10x+16y=22

---------------------

10x-16y=-22

-10x+16y=22

---------------------

0+0=0

infinitely many solutions

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:

$\hat a = max(X_i)$

$E(a) - a= 0$ 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

In a box there are 21 identical pieces that have been enumerated using odd numbers from 5 to 45. How many tiles should be drawn

Orlov [11]

These are the only combinations of exactly 3 tiles that add to 33.

5,7,21

5,9,19

5,11,17

5,13,15

7,9,17

7,11,15

9,11,13

All tiles with numbers above 21 do not help you. 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45. There are 12 of them.

Every three-number combination must have one of the numbers 5, 7, or 9, so if the six numbers from 13 to 21 are picked, in addition to the 12 higher numbers mentioned above, you already picked 18 tiles, and you still have no solution. To obtain the solutions 5,7,21; 5,9,19; 7,9,17; he needs two more numbers in addition to the 18 he already has, so he needs 20 tiles in total to be guaranteed three of them add to exactly 33.

Answer: 20 tiles

4 0

3 years ago

There are 5 boys for every 6 girls.How many boys would there be if there were 36 girls?

dimulka [17.4K]

Answer:

Step-by-step explanation:

If there are 5 for every 6, then you divide 36 by 6 and then you get 6

Once you get that, multiply 5 by 6 and get 30

4 0

3 years ago

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