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Liono4ka [1.6K]
3 years ago
11

Solve equestion give i give you 20 points​

Mathematics
2 answers:
WARRIOR [948]3 years ago
7 0

Answer:

i cant see everything

Step-by-step explanation:

andre [41]3 years ago
3 0

Answer:

Is that the whole problem

Step-by-step explanation:

or is there more we need to see

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Answer with detail. Please show work for all questions (worth 100 points plus ill mark brainliest) ^-^ Thanks!
polet [3.4K]

Answer:

see below

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h

We can find the radius from the diameter

r = d/2 = 34/2 = 17

V = (3.14) (17)^2 * 27

V =24501.42 m^3

The surface of a cone is found by

SA = pi r^2 + pi r  l where l is the slant height

We can find the radius from the diameter

r = 8/2 = 4

SA =(3.14) * 4^2 + 3.14 * 4*7

      =50.24+87.92

        138.16 in^2

The lateral surface area of a pyramid is found by taking the area of the sides

LSA = 4 * the area of a triangle

          we multiply by 4 since there are 4 sides

        = 4 * 1/2 b*h  where b is the base of the triangle and h is the height of the triangle.  (in this case it would be the slant height)

           = 2 * 8 *22

          =352 m^2

7 0
3 years ago
Read 2 more answers
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
ASAP!!!!!!!!!!!!!!!!!!!
timama [110]

Answer:

Step-by-step explanation:

BC = 60

CA = 45

AB = sqrt(45^2 + 60^2)

AB = sqrt(2025 + 3600)

AB = sqrt(5625)

AB = 75

7 0
3 years ago
A customer went to a garden shop and bought some potting soil for $17.50 and 4 shrubs. The total bill was $53.50. Choose the equ
olga2289 [7]
4x + 17.50 = 53.50

the prices of each shrub would be 9 dollars each
6 0
3 years ago
Evaluate 3.6x + 4.5y when x=3 and y=7
Ilya [14]
The answer is 42.3


Step by Step:

Let our variable be n, standing for number.

3.6(x) + 4.5(y) = n
3.6(3) + 4.5(7) = n
10.8 + 31.5 = n
n = 42.3

5 0
3 years ago
Read 2 more answers
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