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Alexxandr [17]
3 years ago
10

A jar contains 7 blue cubes, 4 blue spheres, 5 green cubes, and 6 green spheres. If you select an object at random, what is the

probability that the object is green or a cube? A. 23/22. B. 5/22. C. 9/11. D. 4
Mathematics
1 answer:
tia_tia [17]3 years ago
8 0
The answer is  A. 23/22.

The addition rule is used to calculate the probability of one of the events from multiple pathways. If you want that only one of the events happens (there is OR in a sentence), you will use the addition rule. So, let's just add the probability that the object is green and the probability that the object is a cube.
There are in total 22 objects:
<span>7 blue cubes + 4 blue spheres + 5 green cubes + 6 green spheres = 22 objects.
</span>The probability that the object is green is 11/22 (because there are 11 objects
that are green (5 cubes and 6 spheres) of in total 22 objects).
The probability that the object is a cube is 12/22 (because there are 12 objects that are a cube (7 blue and 5 green) of in total 22 objects).

Thus, the probability that the object is green or a cube is:
\frac{11}{12} + \frac{12}{12} = \frac{11+12}{12} = \frac{23}{12}

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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.

3 0
4 years ago
Please help me with number five write an inequality thank you :)
Igoryamba

Esecially the problem is 3x +x < 160

To find this out we first need to combine like terms

4x<160

Then get rid of the coefficient. In this problem, we divide by 4

x<40

So Rachel has to be less then 40, and Stephanie has to be less than 120.


Any questions?

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4 years ago
At dan's school,6 out of every 10 kids know how to swim. If there are 350 kids in his school,how many know how to swim?
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Answer:

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Step-by-step explanation:

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