Think the quality of it is based off of the amount of time some one worked on something. 2. If all of the data values in the class were compiled in a giant histogram, and a curve was drawn over the graph, where should the highest point on the curve occur? It should be were the mode is because there is the most of the mode in the data set. 3. What phase(s) of a design process might statistics be most applicable? The revising stage. 4. Other than dial calipers, what kinds of tools are used to precisely measure objects for purposes of quality control? A ruler/ measuring tape or you can use a scale and you could measure well.
A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
5 * 2 = 10
3^10/3^-2, same base, add
10 - 2 = 8
Solution: 10^8
Hi, this should be correct. Hope this helps :)
Answer: 5,760
Step-by-step explanation:
It was given that each dog is distinct from other dogs and each cat is distinct from other cats, Also from the hint given, the First position is for the dogs. Let D represent the dogs and C represent the Cat , then we have
D C D C D C D C D D or
D D C D C D C D C D or
D C DD C D C D C D or
D C D C D D C D C D or
D C D C D C DD C D
Each of the arrangements above could be done in
2 x 4! x 4! ( it is constant that D is starting , so I am only left with the arrangement of the remaining 5 D's , out of the remaining 5 D's it is also constant that tow of them will be together and this could be done in 2 ways, so I have 4! left for the D's and 4! also for the C's )
= 2 x 24 x 24
= 1 , 152
The total arrangement = 5 x 1 , 152
= 5,7 60 ways
