Answer:
Assuming the problem asks to rotate counterclockwise: E' (0,0) F' (0,5) G' (-4,5) H' (-4,0)
If it says rotate clockwise then your points will be: E' (0,0) F' (0,-5) G' (4,-5) H' (4,0)
Step-by-step explanation:
Translate the points first by subtracting 3 from the x-coordinates of each, then subtracting 3 from the y-coordinates of each. Next, to rotate them counterclockwise about the origin 90 degrees, switch the x and y coordinates and the sign of the resulting x-coordinate. To rotate clockwise 90 degrees switch the x and y still but change the sign of the resulting y coordinate.
Part A
There are 6 bijections from {1,2,3} to {a,b,c}. This is effectively the same as asking the question "how many ways are there to arrange {a,b,c} where order matters?" We use a factorial to answer this question.
3 factorial = 3! = 3*2*1 = 6
You can also use a permutation, which is composed of factorials, to get the same answer.
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Part B
There are no bijections from {1,2,3} to {a,b,c,d}. Why is this? Because a bijection has two properties: it must be one-to-one, and it must be onto. The term "onto" in mathematics means "every value in the range is targeted". In the case of the range {a,b,c,d} it is not possible for each value to show up. This is because there are only three items in the domain {1,2,3}. You'll always be one letter short.
As you can probably guess, a bijection is only possible if and only if n(D) = n(R), where D and R are the domain and range respectively. The notation n(D) represents the count or number of items in set D.
Firstly, put Derek and Paul's collection together:
1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910, 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920
1950+1952+1908+1902+1955+1954+ 1901+1910+1929+1935+1928+1930+ 1925+ 1932+1933+1920
=30864
=30864÷16
=1929
∴the mean is 1929.
Median:
1901, 1902, 1908, 1910, 1920, 1925, 1928, 1929, 1930, 1932, 1933, 1935,
1950, 1952, 1954, 1955
1930÷1929
=1.0005184
= Median
Range:
1955-1901
=54
∴the range is 54.
Answer:
43
Step-by-step explanation:
