Find the area of a floor that measures 12 1/2 feet by 15 feet.
2 answers:
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Answer:
C . 0.76
Step-by-step explanation:
We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:
Jan 40.55 Feb 41.51
March 42.01 Apr 38.76
May 36.32 June 34.28
July 35.38 Aug 37.48
Sept 40.87 Oct 48.32
Nov 41.59 Dec 42.71
We have to find the difference between the mean of the data , including the outlier and excluding the outlier
Outlier: That observation which is different from other observations.
The outlier in the given observations is 48.32 because is different from other observations.
Mean of the data including the outlier
Mean =
Mean=
Mean=
Mean=39.982
Mean of the data excluding the outlier
Mean=
Mean=39.224
Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758
Difference between mean of the data including the outlier and excluding the outlier=0.76
Answer Answer:
Step-by-step explanation:
We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:
Jan 40.55 Feb 41.51
March 42.01 Apr 38.76
May 36.32 June 34.28
July 35.38 Aug 37.48
Sept 40.87 Oct 48.32
Nov 41.59 Dec 42.71
We have to find the difference between the mean of the data , including the outlier and excluding the outlier
Outlier: That observation which is different from other observations.
The outlier in the given observations is 48.32 because is different from other observations.
Mean of the data including the outlier
Mean =
Mean=
Mean=
Mean=39.982
Mean of the data excluding the outlier
Mean=
Mean=39.224
Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758
Difference between mean of the data including the outlier and excluding the outlier=0.76
Answer :C . 0.76
Answer:
first exercise
V = 16 π
second exercise
V= 68π/15
Step-by-step explanation:
Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4
<em>vide</em> picture 1
The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence, this new red equation is defined by x = (y − 1)2.
Now, we have two equations.
x = (y − 5)2
xm = (y − 1)2 (Rotated graph in red on the figure)
The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.
Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.
(1) Vt = V - Ve
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
f(y) represents the radius of the infinitesimal cylinder.
Replacing (2) in (1), we have
V = 16 π
----
Second part
Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.
<em>vide</em> picture 2
As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.
The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.
Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:
Notice that in the equation above, x and y are switched to facilitate the calculation. f(x) is the radius of the infinitesimal cylinder
Having this in mind, the infinitesimal radius of equation (3) is defined by f(x) + radius of the revolution, which is 7. The volume seeked is the volume defined by the y = x2 minus the volume defined by x=y2. As follows:
V= 68π/15
