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See attached for a sketch of some of the cross sections.
Each cross section has area equal to the square of the side length, which in turn is the vertical distance between the curve y = √(x + 1) and the x-axis (i.e. the distance between them that is parallel to the y-axis). This distance will be √(x + 1).
If the thickness of each cross section is ∆x, then the volume of each cross section is
∆V = (√(x + 1))² ∆x = (x + 1) ∆x
As we let ∆x approach 0 and take infinitely many such cross sections, the total volume of the solid is given by the definite integral,
Answer:
Average size of the lots = ⅓ acre
Step-by-step explanation:
The question incomplete without specifying what we are to determine.
Question:A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X's. Organize the information in a line plot. What is the average size of the lots? _________ acre
Help anyone?
Solution:
We are asked to organize the information in a line plot. See attachment for the line plot.
Given: 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X'sIn no particular order, the sizes of the lots are:1/6, 1/6, 1/6, 1/6, 1/6, 1/6, 1/3, 1/3, 1/3, 1/2, 1/2, 1/2, 1/2, 1/2 and 1/2 acre.
Let's count the number of lot for each size given.
For 1/6: there are 6 X's on the line plot of 1/6 number of lot for 1/6
= the lot × number of times it occurs
= (1/6) × 6 = 6/6 = 1 acre
For 1/3: there are 3 X's on the line plot of 1/3 number of lot for 1/3 = the lot × number of times it occurs
= (1/3) × 3 = 3/3 = 1 acre
For 1/2: there are 6 X's on the line plot of 1/2 number of lot for 1/2
= the lot × number of times it occurs
= (1/2) × 6 = 6/2= 3 acres
To find average size of the lots, we would sum all lot for each given size then divide by the total number of lots given.Sum of all lot for each given size = 1+1+3
Sum of all lot for each given size = 5
The total number of lots given = 15
Average size of the lots = 5/15 = 1/3
Average size of the lots = ⅓ acre
