Answer:
a) 
b) 
And we have the following table:
X | 0 | 1 | 2
P(X) | 0.25 | 0.5 | 0.25
Step-by-step explanation:
Let's define first some notation
H= represent a head for the coin tossed
T= represent tails for the coin tossed
We are going to toss a coin 2 times so then the size of the sample size is 
a. What is the sample space for this chance experiment?
The sampling space on this case is given by:
b. For this chance experiment, give the probability distribution for the random variable of the total number of heads observed.
The possible values for the number of heads are X=0,1,2. If we assume a fair coin then the probability of obtain heads is the same probability of obtain tails and we can find the distribution like this:
And we have the following table:
X | 0 | 1 | 2
P(X) | 0.25 | 0.5 | 0.25
Step-by-step explanation:
There is no diagram so I cannot understand how to answer
<span>Another answer:
According to the figure, the coordinates of JKLM are, J(-7,-2), K(-4,-2), L(-2,-5) and M(-9,-5)
JKLM is translated to J'K'L'M' by means of the rule (x+8, y-3), we should know that (x, y) are coordinates of the pre image, to find the coordinates of each image we have (x', y') such that
x' =x+8, and y' =y-3. Therefore, the coordinates of L' can be found by L'(-2+8, -5-3)=(6, -8)
the final answer is L'(6, -8).</span>
AFB, the sides are similar
Answer:
<h3>120 different ways</h3>
Step-by-step explanation:
Using the combination formula as shown;
nCr = n!/(n-r)!r!
If a stack of 10 different cards are shuffled and spread out face down. If 3 cards are turned face up, the number of different ways 3cards combinations are possible is expressed as;
10C3 = 10!/(10-3)!3!
10C3 = 10!/(7)!3!
10C3 = 10*9*8*7!/7!*3*2
10C3 = 10*9*8/3*2
10C3 = 720/6
10C3 = 120 different ways
Hence there are 120 different card combinations
