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Anestetic [448]
3 years ago
7

Figure ABCD is a trapezoid with point A (0, -4). what rule would rotate the figure 270 clockwise?

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
6 0
The rule to rotate the figure is (-y,x) making point A (4,0).
You might be interested in
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
PLEASE HELP ME ITS DUE TOMORROW!! THANKS!
leonid [27]
PART A
Change the fractions into improper fractions
pablo - rosa = 4 4/9 - 3 5/12
pablo - rosa = 40/9 - 41/12

Equalize the denominator of the fractions
I equalize them to 36. If the denominator 9 is multiplied by 4, so is the numerator. If the denominator 12 is multiplied by 3, so is the numerator.
pablo - rosa = 40/9 - 41/12
pablo - rosa = (40 × 4)/(9 × 4) - (41 × 3)/(12 × 3)
pablo - rosa = 160/36 - 123/36
pablo - rosa = 37/36

Change it to mixed fraction
pablo - rosa = 37/36
pablo - rosa = 1 1/36

Pablo has 1 1/36 quarts more than Rosa

PART B
Calculate the iced tea Pablo gave to Rosa
Change into proper fraction/improper fraction
iced tea given = 15% × 4 4/9
iced tea given = 15/100 × 40/9
iced tea given = 600/900
iced tea given = 2/3

Calculate Pablo's iced tea after giving
Pablo's = 40/9 - 2/3
Pablo's = 40/9 - (2 × 3)/(3×3)
Pablo's = 40/9 - 6/9
Pablo's = 34/9
Pablo's = 3 7/9

Calculate Rosa's iced tea
Rosa's = 41/12 + 2/3
Rosa's = 41/12 + (2 × 4)/(3 × 4)
Rosa's = 41/12 + 8/12
Rosa's = 49/12
Rosa's = 4 1/12

Pablo has 3 7/9 quarts and Rosa has 4 1/12 quarts
6 0
3 years ago
What would this One be??
inn [45]

Answer:

the pre-image and the image are congruent

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A teacher has 36 sheets of construction paper and 78 markers. She wants to pack them in boxes with the same number of sheets and
ArbitrLikvidat [17]

Answer:

6 boxes.

Step-by-step explanation:

We have been given that a teacher has 36 sheets of construction paper and 78 markers.

As teacher wants to pack the same number of sheets and markers in each box, So we will have to find the greatest common factor (GCF) of 36 and 78.  

Factors of 36 are: 1, 2, 3, 4, 6, 6, 18 ,12, 18, 36

Factors of 78 are: 1, 2, 3, 26, 39, 78.

We can see that GCF of 36 and 78 is 6. The number of sheets of construction paper in each box will be 6 (6*6=36) and number of markers in each box will be 13 (13*6=78).

Therefore, the greatest number of boxes the teacher can use to pack the sheets of construction paper and markers is 6.

5 0
4 years ago
Could I get help with figuring out 6 √-63
barxatty [35]

Answer:

undefined

Step-by-step explanation:

The square root of a negative number is undefined

3 0
3 years ago
Read 2 more answers
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