All categories

3 years ago

Question 19 Unsaved

Mathematics

1 answer:

nata0808 [166]3 years ago

8 0

A response variable you are trying to measure once you've done the experiment.

What have you done? You've measured what sunglasses will do to affect eyesight. You have used a 10 week period to measure their response. You don't really care what the response is -- only if there is one.

10 people and 10 weeks don't change. Those two are not a choice for an answer. Now you come to the crunch of the question. Is it the sunglasses you are trying to measure or is it what your eyes have done because you were wearing sun glasses.

So we need an answer. If you say sun glasses, how did you measure what they did. THAT's the answer. The answer is not the sunglasses.

You might be interested in

Two independent simple random samples are taken to test the difference between the means of two populations whose variances are

My name is Ann [436]

I think it’s 72 degrees of freedom

3 0

3 years ago

HURRY! I need help on my exam. (10pts!)

Anna [14]

Answer:

oof

Step-by-step explanation:o o o f

7 0

3 years ago

I needddd helppp pleaseeeeee

UNO [17]

Answer:

has to be letter b because they connect and then separate opposite sides

3 0

3 years ago

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali

Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

$W =$ Wednesday

$T =$ Thursday

$F =$ Friday

$S =$ Saturday

Their corresponding probabilities are

$P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1$

Since $Y$ = number of days beyond Wednesday that it takes for both magazines to arrive(so possible $Y$ values are 0, 1, 2 or 3)

The possible number of outcomes are therefore $4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)$

The values associated for each of the outcomes are as follows:

$Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3$

The probability mass function of $Y$ is,

$P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19$

7 0

3 years ago

Plz help! ill mark brainiest

MrRa [10]

Answer:

part a) linear, linear

part b) function #1

3 0

3 years ago