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natulia [17]
3 years ago
15

Determine if \sqrt(9) is rational or irrational and give a reason for your answer

Mathematics
1 answer:
madreJ [45]3 years ago
3 0

Answer:

/sqrt(9) would be rational. the square root of 9 is 3, 3x3 is 9. a irrational number cannot be written as a simple fraction.

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Marjorie has 28 feet of trim to use as edging on a rectangular blanket she wants to make. What is the length and width of two bl
zloy xaker [14]
7 by 2 for each blanket. If you split 28 into 2 sheets, its 14, sooo.
5 0
3 years ago
I can’t find the fraction for 0.45
Andre45 [30]

Answer: 45/100

Step-by-step explanation:  .45 as a fraction is 45/100

4 0
3 years ago
Read 2 more answers
Sociologists say the percentage of the classic music lovers is about 5%. From the survey, we know that only 83 out of 2000 peopl
disa [49]

Answer:

p value from chi square = 0.081

Step-by-step explanation:

Given that sociologists say the percentage of the classic music lovers is about 5%. From the survey, we know that only 83 out of 2000 people listen to the classic music. Let denote the proportion of the classic music lovers. The hypotheses are H0: p=0.05 vs Ha: p≠0.05

p value = 0.09

Instead of this if we do chi square

H0: Proportion of music lovers is 5% and non music lovers is 95%

Ha: Proportions are not as per the estimate

(Two tailed chi square test)

Observed       83           1917       2000

Expected        100          1900     2000

O-E                  -17              17          0

chi square

(O-E)^2/E        2.89         0.1521     3.0421

df =1

p value from chi square = 0.081

Though p value is not exactly the same, the conclusion in both tests are the same

5 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
Simplify the expression √6 (√33=7)
marin [14]
14.07 is the answer to the question above
7 0
3 years ago
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