Considering that the blue cup has to be on one end, we consider the number of arrangements possible with the other 4 cups, then multiply by 2(considering the blue on each end).

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Considering green and yellow before red and purple, there are $2 \times 2 = 4$ outcomes(G-Y-R-P, G-Y-P-R, Y-G-R-P and Y-G-P-R).
Another two possible outcomes are yellow-purple and green-red, or vice-versa, thus 4 + 2 = 6 outcomes.
To account for the blue on the end, multiplying by 2. $2 \times 6 = 12$
There are 12 possible arrangements.

A similar problem is given at brainly.com/question/24617788

6 0

3 years ago

Solve the equation. X- 8 = 12

beks73 [17]

Answer:

4 0

3 years ago

Read 2 more answers

The scientist performs additional analyses and observes that the number of major earthquakes does appear to be decreasing but wo

kifflom [539]

Answer:

Step-by-step explanation:

Hello!

A regression model was determined in order to predict the number of earthquakes above magnitude 7.0 regarding the year.

^Y= 164.67 - 0.07Xi

Y: earthquake above magnitude 7.0

X: year

The researcher wants to test the claim that the regression is statistically significant, i.e. if the year is a good predictor of the number of earthquakes with magnitude above 7.0 If he is correct, you'd expect the slope to be different from zero: β ≠ 0, if the claim is not correct, then the slope will be equal to zero: β = 0

The hypotheses are:

H₀: β = 0

H₁: β ≠ 0

α: 0.05

The statistic for this test is a student's t: $t= \frac{b - \beta }{Sb} ~~t_{n-2}$

The calculated value is in the regression output $t_{H_0}= -3.82$

This test is two-tailed, meaning that the rejection region is divided in two and you'll reject the null hypothesis to small values of t or to high values of t, the p-value for this test will also be divided in two.

The p-value is the probability of obtaining a value as extreme as the one calculated under the null hypothesis:

p-value: $P(t_{n-2}\leq -3.82) + P(t_{n-2}\geq 3.82)$

As you can see to calculate it you need the information of the sample size to determine the degrees of freedom of the distribution.

If you want to use the rejection region approach, the sample size is also needed to determine the critical values.

But since this test is two tailed at α: 0.05 and there was a confidence interval with confidence level 0.95 (which is complementary to the level of significance) you can use it to decide whether to reject the null hypothesis.

Using the CI, the decision rule is as follows:

If the CI includes the "zero", do not reject the null hypothesis.

If the CI doesn't include the "zero", reject the null hypothesis.

The calculated interval for the slope is: [-0.11; -0.04]

As you can see, both limits of the interval are negative and do not include the zero, so the decision is to reject the null hypothesis.

At a 5% significance level, you can conclude that the relationship between the year and the number of earthquakes above magnitude 7.0 is statistically significant.

I hope this helps!

(full output in attachment)