Yes! For example, 10 and 20 have 1, 2, 5, & 10 in common ^.^
Answer:
Step-by-step explanation:
hey i got the answer straight from my teach but ony a and c if you have the answer to b let me know : ( A) To determine the mean absolute deviation, first calculate the mean. Find the sum of the values: 8 plus 6 plus 9 plus 6 plus 14 plus 9 plus 5 plus 7 equals 64. Then divide the sum by the number of values: 64 divided by 8 equals 8. The mean is 8.
Next, find the distance of each value from the mean :
|8-8|=0 |6-8|=2 |9-8|=1 |6-8|=2 |14-8|=6 |9-8|=1 |5-8|=3 |7-8|=1
0 plus 2 plus 1 plus 2 plus 6 plus 1 plus 3 plus 1 equals 16. Then divide the sum of absolute deviations by the number of data values: 16 divided by 8 equals 2. The MAD of the data is 2.
C. Since the MAD for the vegetable plant data is so much higher than the MAD for the flower data, this indicates that the heights of the vegetable plants vary more than the heights of the flowers ALSO do you go to k12
Answer:
Improper fraction: 39/34
Proper fraction: 1 5/34
Decimal: 1.147 (rounded to the nearest thousandths).
Step-by-step explanation:
Solve each expression given, if it has the same result, then the expressions is equivalent. Solve the given expression:
4 7/8 ÷ 4 1/4
First, change all fractions into improper fractions:
4 7/8 = (4 * 8)/8 + 7/8 = 32/8 + 7/8 = 39/8
4 1/4 = (4 * 4)/4 + 1/4 = 16/4 + 1/4 = 17/4
Next, find common denominators. Note that what you do to the denominator, you must do the numerator:
(17/4) * (2/2) = (34/8)
39/8 ÷ 17/4
Solve. First, change the division sign into multiplication, and then flip the second fraction:
39/8 ÷ 17/4 = 39/8 x 4/17
Multiply straight across:
39/8 x 4/17 = (39 x 4)/(8 x 17) = 156/136 = 39/34 (simplified), or 1 5/34.
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The solution to the problem is as follows:
<span>7(4)/4(4) = 28/16
28+16= 44
</span>
Therefore, the size of the graduating class would be 44 students in all.
I hope my answer has come to your help. God bless and have a nice day ahead!
Answer:
45
Step-by-step explanation:
Let the three quantities be x, y and z.
