B) Histograms can be used to exhibit the shape of distributions.
You would have one tenths (1/10) of a chance to get a 4 on the first go. you would have four ninths (4/9) of a chance to get a number less than 5 out of the bag on the second go.
the overall probability would be two forty-fifths (2/45)
(m+2)(m+3)= (m+2)(m-2)
⇒ m^2+ 3m+ 2m+ 6= m^2 -4
⇒ 5m+ 6= -4 (m^2 on both sides cancels out)
⇒ 5m= -4-6
⇒ 5m= -10
⇒ m= -10/5
⇒ m= -2
The final answer is m=-2~
