Plz help with 9 and 10
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5 and 6 and 3 and 10, the pairs of numbers that have a least common multiple of 30.
<u>Step-by-step explanation:</u>
Case 1: LCM (3, 6)
Prime factorization of 3:
Prime factorization of 6:
Using all prime numbers found as often as each occurs most often we take
Therefore LCM (3, 6) = 6.
Case 2: LCM (5, 6)
Prime factorization of 5:
Prime factorization of 6:
Using all prime numbers found as often as each occurs most often we take
Therefore LCM (5, 6) = 30
Case 3: LCM (3, 10)
Prime factorization of 3:
Prime factorization of 10:
Using all prime numbers found as often as each occurs most often we take
Therefore LCM (3, 10) = 30
Case 4: LCM (5, 10)
Prime factorization of 5:
Prime factorization of 10:
Using all prime numbers found as often as each occurs most often we take
Therefore LCM (5, 10) = 10
The critical values corresponding to a 0.01 significance level used to test the null hypothesis of ρs = 0 is (a) -0.881 and 0.881
<h3>How to determine the critical values corresponding to a 0.01 significance level?</h3>The scatter plot of the election is added as an attachment
From the scatter plot, we have the following highlights
- Number of paired observations, n = 8
- Significance level = 0.01
Start by calculating the degrees of freedom (df) using
df =n - 2
Substitute the known values in the above equation
df = 8 - 2
Evaluate the difference
df = 6
Using the critical value table;
At a degree of freedom of 6 and significance level of 0.01, the critical value is
z = 0.834
From the list of given options, 0.834 is between -0.881 and 0.881
Hence, the critical values corresponding to a 0.01 significance level used to test the null hypothesis of ρs = 0 is (a) -0.881 and 0.881
Read more about null hypothesis at
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