<span>Table:
Class Boundaries Frequency
5-10 8
10-15 9
15-20 15
20-25 10
25-30 8
30-35 6
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total 56
Average =
[5+10]/2*8+[10+15]/2*9+[15+20]/2*15+[20+25]/2*10+[25+30]/2*8+[30+35]/2*6
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</span> 56
That is 1075 / 56 = 19.2
Answer: 19
Answer:
Change in percent = 56.75 (Approx.)
Step-by-step explanation:
Given:
Score in first game = 37 points
Score in second game = 58 points
Find:
Change in percent
Computation:
Change in percent = [(Score in second game - Score in first game)/Score in first game]100
Change in percent = [(58-37)/37]100
Change in percent = [(21)/37]100
Change in percent = 56.75 (Approx.)
The answer is <u>0.001213 mi</u>
Answer:
(25.732,30.868)
Step-by-step explanation:
Given that in a random sample of 42 people, the mean body mass index (BMI) was 28.3 and the standard deviation was 6.09.
Since only sample std deviation is known we can use only t distribution
Std error = 

t critical for 99% two tailed 
Margin of error
Confidence interval lower bound = 
Upper bound = 
This is the answer, hope this helps you