Gracie, Mary, and Nancy each have a small collection of seashells. Gracie has 5 more than times the number of shells Mary has. N
ancy has 1 more than times the number of shells Mary has. Gracie and Nancy have the same number of shells. If x is the number of shells Mary has, identify the equation that represents this situation and identify its solution.
2 answers:
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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
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