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1 answer:
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Answer:
a. See attachments
B. 163630.2ft
C. 147056.70ft
D. 49205.4ft
Step-by-step explanation:
Check attachments for details
Angle a.. Angle of dep to beginning of runway
Angle b... Angle of del to end of runway
AD... Altitude of plane
DB..... Ground distance before touch down
AC... Air distance travelled until touch down on the near end of runway
Answer:
Question 1) 17.4
Question 2) 1.7
Question 3) 2.8
Question 4) 8.6
Step-by-step explanation:
Mean absolute deviation is the average distance between the data points from the set to the mean point of the data set. It shows the variability in data or how much the data points are spread.
method to calculate Mean absolute deviation:
a. calculate mean
b. calculate absolute deviation of each data point
c. add all the deviations
d. divide absolute deviation by number of points
mean absolute deviation = ∑l-xl / n
The given problem has four sub-parts
Solution 1:
Mean= (78+99+90+80+55+56+102+88+60+42)/10
= 750/10
= 75
Mean absolute deviation= ( I 78-75 I + I 99-75 I+I 90-75 I + I 80-75 I +
I 55-75 I + I 56-75 I + I 102-75 I + I 88- 75 I +
I 60-75 I + I 42-75 I) /10
= (174)/10
= 17.4
Solution 2:
Mean = (10+13+7+12+9+8+12+10+11+13)/10
= (105)/10
= 10.5
Mean absolute deviation = ( I 10-10.5 I + I 13-10.5 I + I 7-10.5 I + I 12-10.5 I +
I 9-10.5 I + I 8-10.5 I + I 12-10.5 I + I 10-10.5 I +
I 11-10.5 I + I 3-10.5 I) /10
= 17/10
= 1.7
Solution 3:
Mean= (1+7+10+5+3+3+6+12+9+4)/10
= 60/10
= 6
Mean absolute deviation = ( I 1-6 I + I 7-6 I + I 10-6 I + I 5-6 I + I 3-6 I +
I 3-6 I + I 6-6 I + I 12- 6 I + I 9-6 I + I 4-6 I) /10
= (28)/10
= 2.8
Solution 4:
Mean= (30+46+25+45+18+25+15+32+40+24)/10
= 300/10
= 30
Mean absolute deviation = ( I 30-30 I + I 46-30 I + I 25-30 I + I 45-30 I +
I 18-30 I + I 25-30 I + I 15-30 I + I 32-30 I +
I 40-30 I + I 24-30 I) /10
= (86)/10
= 8.6
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