HELPPPPPPP!!!!!! i need this today

Uh need help

deff fn [24]

Answer:

$48

Step-by-step explanation:

price of jacket : original price

3 : 4

If 64 = 4, we can find out 1.

64 / 4 = 16

16 x 3 = 48

16 x 4 = 64

48 : 64

sale price = $48

Hope this helps!

- profparis

3 0

1 year ago

Read 2 more answers

In 1900, there was an Olympic underwater swimming event. The score was calculated by giving one point for each second the swimme

mash [69]

68.4*1= 68.4 60.2*2= 120.4 68.4+120.4= 188.8

4 0

3 years ago

Read 2 more answers

Alvin is 15 years younger than Elga. The sum of their ages is 25. What is Elga’s age

Fynjy0 [20]

Elga is 20 years old and Alvin would be 5. Hope this helped.

4 0

3 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

3 years ago

Can someone plz answer B and C

Shkiper50 [21]

B.
When two lines intersect they form two pairs of opposite angles.
Vertical angles are always congruent, which means that they are equal.
so m<7 = m<8 or m<3y + 19 = m<5y-29

C. solve for y and measure <7 and <8

3y + 19 = 5y-29
5y - 3y = 19+29
2y = 48
y = 24

so
5y-29 = 5(24) - 29 = 91
3y + 19 = 3(24) + 19 = 91

so <7 = 180 - 91 = 89
<7 = <8 = 89

4 0

3 years ago