a. There are four 5s that can be drawn, and 
ways of drawing any three of them. There are 
ways of drawing any three cards from the deck. So the probability of drawing three 5s is
In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of 
of getting drawn.
b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability 
of being drawn.
c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of 
.
- - -
Note: 
is the binomial coefficient,
Answer:
0.209205
Step-by-step explanation:
I put it in the calculator and got this...
Tan x=opposite/adjacent
cot x=adjacent/opposite
so you need to inverse the fraction
tan x=.78 as a fraction is 78/100
-> inversed= 100/78=1.28=cot x
so it is the second option
Think of the "of" as a multiplication sign:
(.15)(12)= 1.8 meters
There you go! hope this helps with this problem and future ones.
Answer:
Step-by-step explanation:
Given
Required [Missing from the question]
G(T(x))
We have:
This implies that:
Substitute:
![G(T(x)) = 3[9(x + 6.9)]](https://tex.z-dn.net/?f=G%28T%28x%29%29%20%3D%203%5B9%28x%20%2B%206.9%29%5D)
Open bracket
