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3 years ago

In how many different ways can the letters of the word mathematics be arranged?

2 answers:

4 0

Treat the vowels like they are 1 block, as if they are one letter. But in this vowel block there are 4 vowels, 2 of which are the same. The number of ways of arranging these 4 vowels is then 4!2!4!2!.

Now, treating this block aa one letter, we have 7 consonants and 1 block, so there are 8 'letters' to arrange with two M's and two T's. So we have to divide out by 2! for each. So we get: 8!2!2!8!2!2!.

Finally we multiply this by the number of ways the vowels can be arranged to get:

4!2!8!2!2!

qwelly [4]3 years ago

4 0

2 i believe would be the answer

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Step-by-step explanation:

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3 years ago

The value of the coefficient of correlation ( r) a. can never be equal to the value of the coefficient of determination (r2). b.

gulaghasi [49]

Answer:

d. can be equal to the value of the coefficient of determination (r2).

True on the special case when r =1 we have that $r^2 = 1$

Step-by-step explanation:

We need to remember that the correlation coefficient is a measure to analyze the goodness of fit for a model and is given by:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The determination coefficient is given by $R= r^2$

Let's analyze one by one the possible options:

a. can never be equal to the value of the coefficient of determination (r2).

False if r = 1 then $r^2 = 1$

b. is always larger than the value of the coefficient of determination (r2).

False not always if r= 1 we have that $r^2 =1$ and we don't satisfy the condition

c. is always smaller than the value of the coefficient of determination (r2).

False again if r =1 then we have $r^2 = 1$ and we don't satisfy the condition

d. can be equal to the value of the coefficient of determination (r2).

True on the special case when r =1 we have that $r^2 = 1$

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