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3 years ago

Simplify (^3)x4 Easy 30 points

Mathematics

2 answers:

aleksklad [387]3 years ago

5 0

Answer:

$=81$

Step-by-step explanation:

$\left(-3\right)^4$

Use exponent rule $\left(-a\right)^n=a^n$ :

$=3^4$

$=81$

nadezda [96]3 years ago

3 0

The answer is 81 because when you multiply 3 by itself four times you get 81

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students must In order for a class trip to be scheduled, at least sign up. So far, 23 students have signed up. Write and solve a

rusak2 [61]

I have added a screenshot with the complete question.

Answer:
At least 7 more students need to sign up for the field trip to be scheduled

Explanation:
Assume that the least number of students remaining to register for the trip to be scheduled is x.

total number of students = students who already registered + x

We are given that, the least number of students required for this trip is 30. This means that:
total number of students ≥ 30
Now, we know that 23 students already registered.

The inequality now is:
x + 23 ≥ 30
x ≥ 30 - 23
x ≥ 7

This means that at least 7 more students must register for the trip to be scheduled.

Hope this helps :)

7 0

3 years ago

When an ANOVA comparing the means of 3 groups indicates that at least one group is different from the others, a common follow-up

lorasvet [3.4K]

Answer:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C$

Since we reject the null hypothesis we want to see which method it's the best to determine which group(s) is (are) different is pairwise two-sample t-tests each assessed using.

And on this case the best option is:

ii) a Bonferonni-corrected alpha level of 0.0167 to control the type I error rate for the overall inference to 5% .

The reason is because the Bonferroni correction "compensates for that increase by testing each individual hypothesis at a significance level of $\alpha$ who represent the desired overall alpha level and m is the number of hypotheses". For our case m=3 hypotheses with a desired $\alpha = 0.05$ , then the Bonferroni correction would test each individual hypothesis at $0.05/3=0.0167$

One advatange of this method is that "This method not require any assumptions about dependence among the p-values or about how many of the null hypotheses are true" . And is more powerful than the individual paired t tests.

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4 years ago

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MakcuM [25]

Answer:

120

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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