To find the mean (average), u add up all ur numbers, then divide by how many numbers there are.
mean = (5 + 12 + 1 + 5 + 7) / 5 = 30/5 = 6
To find the mode (and there does not have to be one)...u find the number that is used the most. That number would be 5...because it is used twice, whereas, the other numbers are just used once.
Answer:
-5/2, -2, 1.7
Step-by-step explanation:
Answer:
Mr Doyle washed the most.
4/15 still need to be washed.
Step-by-step explanation:
2/5=6/15 Mr Doyle
1/3=5/15 Son
so Mr Doyle washed the most.
Together they washed 11/15 of the laundry so 4/15 is left to wash.
Answer:
-3.1, -3.2, -3.3, -3.4, -3.5, -3.6
Step-by-step explanation:
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
