a metal pile is 500cm long correct to the nearest centimeter the pole is cut into rods each of length 5.8cm correct to the neare
1 answer:
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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
!
Use TrianCal to draw a triangle with phi as Great Piramid (minimum perimeter given 2 equal heights) = maximun stability.
NOTE: Phi=(1+√5)/2≈1.62 and acos(1/Phi)≈51.83º
NOTE: Phi=(1+√5)/2≈1.62 and acos(1/Phi)≈51.83º