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drek231 [11]
4 years ago
7

"calculate the number of permutations possible when using the first five letters of the alphabet to create 3-letter words.

Mathematics
1 answer:
Hitman42 [59]4 years ago
5 0
That is   5! / (5-3)!  =  120 / 2 = 60 answer
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Step-by-step explanation:

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3 years ago
A professor claims that his students' average score on the first exam of the semester is different than the average score on the
Marta_Voda [28]

Answer:

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before (first exam) , y = test value after (second exam)

The system of hypothesis for this case are:

Null hypothesis: \mu_y -\mu_x =0

Alternative hypothesis: \mu_y -\mu_x \neq 0

The first step is calculate the difference d_i=y_i-x_i

The statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

The next step is calculate the degrees of freedom given by:

df=n-1=8-1=7

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

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