In a symmetric histogram, you have the same number of points to the left and to the right of the median. An example of this is the distribution {1,2,3,4,5}. We have 3 as the median and there are two items below the median (1,2) and two items above the median (4,5).
If we place another number into this distribution, say the number 5, then we have {1,2,3,4,5,5} and we no longer have symmetry. We can fix this by adding in 1 to get {1,1,2,3,4,5,5} and now we have symmetry again. Think of it like having a weight scale. If you add a coin on one side, then you have to add the same weight to the other side to keep balance.
Answer:
624
Step-by-step explanation:
find volumes of box and cubes(in inches)
divide box volume by cube volume
Answer:
For (fg)(x)
(fg) (x) = 4x^4 - 8x^3 -11x^2 -3x
With no restrictions on the x
Step-by-step explanation:
To find (fg) (x) = f(x) . g(x)
We need to multiply f(x) with g(x)
(fg) (x) = (2x^2 -5x -3) * (2x^2 + x)
fg(x) = 4x^4 + 2x^3 - 10x^3 - 5x^2 -6x^2 -3x
fg(x) = 4x^4 - 8x^3 -11x^2 -3x
Answer:
Step-by-step explanation:
Let f be a function from N to N.
N_set of all natural numbers
i) one to one but not onto
consider the function
When two numbers have same square we find that the numbers should be the same because they are positive.
So one to one but not onto because consider 3 it does not have square root in N.
ii) Onto but not one to one
Consider
this is onto because every number has a preimage in N.
But not onto because consider 6 and 3, f(6) = 3 and f(3) =3
So not one to one
iii) both onto and one-to-one
f(x) = 
=x+1, x even
This is both one to one and onto since we consider only integers
iv) Neither one to one nor onto
Consider the function
f(x) = 2
This is not onto because 3 cannot have a preimage in N, not one to one because f(1) = f(2) where 1 not equals 2
