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3 years ago

True or False?

1 answer:

7 0

$FALSE!\\The\ segments\ below\ couldn't\ form\ a\ triangle.\\\\a,b,c-lengths\ sides\ of\ a\ triangle,\ then:\\\\a+b > c\\a+c > b\\b+c > c\\\\Here:a=9;\ b=4;\ c=15\\\\a+b=9+4=13 < 15!!!!$

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I need help with this!!!!

Ira Lisetskai [31]

Answer:

D: 2/4

Step-by-step explanation:

Usually when we talk about a point partitioning a segment, we are interested in the ratio of the first segment to the second:

BC : CD = 2 : 2 = 1 : 1

Since this is not an answer choice, we need to "reverse engineer" the answer list to see if we can find an answer that corresponds to a reasonable interpretation of the question.

None of the segments is 3 units long, so neither of answer choices A or B makes any sense.

While segment BD is 4 units long, there is no segment that is 1 unit long, so answer choice C makes no sense, either.

There are segments that are 2 units long and a segment that is 4 units long, so if we interpret the question to be "what is the ratio of BC to BD?" then answer choice D is appropriate.

4 0

3 years ago

An acute angle ? is in a right triangle with cos ? = nine tenths. what is the value of sec ?? square root of nineteen divided by

Anon25 [30]

What...................................numberxs

6 0

3 years ago

Find an equation of the line that has a slope of -8 and y-intercept of 6. Write the answer in the form Y = Mx+B

solong [7]

The answer is y= -8x+6
-8 is the slope and 6 being the y-intercept.

3 0

3 years ago

Read 2 more answers

Given: Lines a and b are parallel and line c is a transversal. Prove: Angle2 is supplementary to Angle8 Horizontal and parallel

kondaur [170]

Answer:

The correct option is;

Corresponding angles theorem

Step-by-step explanation:

Type of lines of lines a and b = Horizontal and parallel lines

The transversal to a and b = Line c

The angles between a and c labelled clockwise from the upper left quarter segment = 1, 2, 4 and 3

The angles between b and c labelled clockwise from the upper left segment = 5, 6, 8 and 7

Therefore, we have;

Statement ${}$ Reason

1. a║b, c is a transversal ${}$ Given

2. ∠6 ≅ ∠2 ${}$ Corresponding angles theorem

3. m∠6 = m∠2 ${}$ Definition of congruent

4. ∠6 is supp. to ∠8 ${}$ Definition of linear pair

5. ∠2 is supp. to ∠8 ${}$ Congruent supplement theorem

Corresponding angles are the angles located in spatially similar or matching corners of two lines that have been crossed by the same transversal. When the two lines having a common transversal are parallel, the corresponding angles will be congruent.

4 0

4 years ago

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

Viefleur [7K]

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

b) For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

Step-by-step explanation:

For this case we assume that we have a random sample given by: $X_1, X_2,....,X_7$ and each $X_i \sim N (\mu, \sigma)$

Part a

In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

5 0

4 years ago