Complete question :
Birth Month Frequency
January-March 67
April-June 56
July-September 30
October-December 37
Answer:
Yes, There is significant evidence to conclude that hockey players' birthdates are not uniformly distributed throughout the year.
Step-by-step explanation:
Observed value, O
Mean value, E
The test statistic :
χ² = (O - E)² / E
E = Σx / n = (67+56+30+37)/4 = 47.5
χ² = ((67-47.5)^2 /47.5) + ((56-47.5)^2 /47.5) + ((30-47.5)^2/47.5) + ((37-47.5)^2/47.5) = 18.295
Degree of freedom = (Number of categories - 1) = 4 - 1 = 3
Using the Pvalue from Chisquare calculator :
χ² (18.295 ; df = 3) = 0.00038
Since the obtained Pvalue is so small ;
P < α ; We reject H0 and conclude that there is significant evidence to suggest that hockey players' birthdates are not uniformly distributed throughout the year.
I think the answer is 210.442
Answer:
(2, 3 )
Step-by-step explanation:
Given the 2 equations
2x - 3y = - 5 → (1)
5x + 4y = 22 → (2) [ rearranged equation ]
Multiplying (1) by 4 and (2) by 3 and adding will eliminate the term in y
8x - 12y = - 20 → (3)
15x + 12y = 66 → (4)
Add (3) and (4) term by term to eliminate y, that is
23x = 46 ( divide both sides by 23 )
x = 2
Substitute x = 2 in either of the 2 equations and evaluate for y
Substituting into (2)
5)2) + 4y = 22
10 + 4y = 22 ( subtract 10 from both sides )
4y = 12 ( divide both sides by 4 )
y = 3
Solution is (2, 3 )
Answer:
You can either do a back to back stem and leaf plot, where you would have double the values. In a normal stem and leaf plot you would just have one set of 3's where you would put all the values that start with 3 in that column. A back to back is the same but instead you would have two 3 values, where anything that is higher than 5 would be in the second value of 3, but anything lower would be in that first value of 3.
