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Hunter-Best [27]

3 years ago

14

First year students in rolled at a college were asked whether they play video games. The responses, classified by whether the st

udents were enrolled in the school of sciences the school of arts are shown in the table.

1 answer:

Irina18 [472]3 years ago

3 0

Where’s the table lol

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4. f(x) = x² – 5 (if x 20) and g(x) = (x + 5

erica [24]

Evaluating functions means to substitute the values at every occurrence of x.

So, we have

$f(x)=x^2-5 \implies f(20)=20^2-5 = 400-5=395$

You didn't specify where we should evaluate g(x).

6 0

4 years ago

The length of the base of an isosceles triangle is x. The length of a leg is 2x - 2. The perimeter of the triangle is 51. Find x

azamat

Answer:

x = 11

Step-by-step explanation:

x + 2(2x-2) = 51

x + 4x - 4 = 51

5x - 4 = 51

5x = 55

x = 11

7 0

3 years ago

Read 2 more answers

Driven any two two number witch is greater the lcm of the number or gcf of the the number?why?

avanturin [10]

Lcm because a multiple is always greater then a factor.

4 0

3 years ago

What is the slope of the line that passes through (-4, 5) and (2, -3)?

ella [17]

Answer:

$m=\frac{-4}{3}$

$y=\frac{-4}{3}x -\frac{1}{3}$

Step-by-step explanation:

Use the Slope formula...

$\frac{-3-5}{2-(-4)}$

$\frac{-8}{6} = \frac{-4}{3}\\\frac{-4}{3} = m$

Plug the slope into the Point-Slope formula...

$y -(5) = \frac{-4}{3}(x-(-4)) \\y+5=\frac{-4}{3}x +\frac{-16}{3}\\y=\frac{-4}{3}x +\frac{-1}{3}$

8 0

3 years ago

Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

a. Then, for $\hat \theta_3$ to be an unbiased estimator ; Then:

$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

Thus; in order to minimize the variance of $\hat \theta_3$ ; then constant a can be determined as :

$V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2$

Using differentiation:

$\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0$

⇒

$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

$\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}$

This implies that

$\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0$

So, $V( \hat \theta_3)$ is minimum when $\hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}$

As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

4 0

4 years ago