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riadik2000 [5.3K]
3 years ago
8

Determine whether the triangles shown can be

Mathematics
2 answers:
lidiya [134]3 years ago
3 0

Answer:

not similar

Step-by-step explanation:

180-56-83=41

Margarita [4]3 years ago
3 0

Answer:

180-56-83=41

Step-by-step explanation:

#LetsStudy

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Find the derivative of the inverse function of the original function f(x)=(x+1)^3-5
____ [38]

Answer:

Step-by-step explanation:

let f(x)=y

y=(x+1)^{3} -5\\flip ~x~and~y\\x=(y+1)^{3}-5\\(y+1)^{3}=x+5\\y=(x+5)^{\frac{1}{3} } -1\\or~f^{-1}(x)=(x+5)^{\frac{1}{3} } -1\\(f^{-1})'(x)=\frac{1}{3} (x+5)^{\frac{-2}{3} }

4 0
3 years ago
Part A- What is the discount rate at the store closing?
Andreas93 [3]
1 question I believe it’s A. 2 question I believe it’s C. 3 question I believe it’s B. Hope this help❤️.
8 0
3 years ago
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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
How do you solve and simplify
DedPeter [7]
F(x) = 1/(x+2)   & g(x) = x/(x-3)


(f(x) + g(x) = 1/(x+2) + x/(x-3). Reduce to same denominator:

1/(x+2) + x/(x-3) =(x-3) + x(x-3)/(x+2).(x-3) ==> (x²+3x-3)/(x+2).(x-3)

7 0
3 years ago
If 2(6 + 3x) + 1 = - 5 + 3(x - 1), what is the value of x?
Stels [109]

Answer: x = -7

Step-by-step explanation:

2(6 + 3x) + 1 = - 5 + 3(x - 1)

(2) (6) + (2) (3x) + 1 = -5 + (3) (x) + (3) (-1)

12 + 6x + 1 = - 5 + 3x + - 1

( 6x ) + ( 12 - 1 ) = ( 3x ) + ( -5 + -3 )

6x + 13 = 3x - 8

6x + 13 - 3x = 3x - 8 - 3x

3x + 13 = -8

3x + 13 - 13 = -8 - 13

3x = -21

3x/3 -21/3

X = -7

5 0
2 years ago
Read 2 more answers
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