Answer:
HHH, W = 3-0 = 3
HHT, W= 2-1=1
HTH, W= 2-1=1
THH, W=2-1 =1
HTT, W= 1-2=-1
THT, W= 1-2=-1
TTH, W=1-2=-1
TTT, W=0-3 = -3
So then the sample space for W is:
![S= [-3,-1,1,3]](https://tex.z-dn.net/?f=%20S%3D%20%5B-3%2C-1%2C1%2C3%5D)
Just 4 possible values from 8 possible combinations for the 3 random tosses
Step-by-step explanation:
For this case we define W as the random variable who represent the number of heads minus the number of tails in three tosses of a coin.
W= # heads- # coins
Since we toss a coin 3 times we have 2*2*2= 8 possible results. We can list the results and the corresponding values for W like this:
HHH, W = 3-0 = 3
HHT, W= 2-1=1
HTH, W= 2-1=1
THH, W=2-1 =1
HTT, W= 1-2=-1
THT, W= 1-2=-1
TTH, W=1-2=-1
TTT, W=0-3 = -3
So then the sample space for W is:
![S= [-3,-1,1,3]](https://tex.z-dn.net/?f=%20S%3D%20%5B-3%2C-1%2C1%2C3%5D)
Just 4 possible values from 8 possible combinations for the 3 random tosses
Answer:

Step-by-step explanation:
According to the trigonometric ratios in aright triangle :

Given: ΔBCA ~ ΔSTR
Since , corresponding angles of two similar triangles are equal.
So, ∠C = ∠T ...(i) [Middle letter]
In triangle STR
...(ii)
From (i) and (ii), we have

Answer: Our required probability is 0.83.
Step-by-step explanation:
Since we have given that
Number of dices = 2
Number of fair dice = 1
Probability of getting a fair dice P(E₁) = 
Number of unfair dice = 1
Probability of getting a unfair dice P(E₂) = 
Probability of getting a 3 for the fair dice P(A|E₁)= 
Probability of getting a 3 for the unfair dice P(A|E₂) = 
So, we need to find the probability that the die he rolled is fair given that the outcome is 3.
So, we will use "Bayes theorem":

Hence, our required probability is 0.83.
The third one: y-1=56(x+8), in the pair -8, 1 the first number always is "x1" and the second number is always "y1".
So the formula is y-y1=slope(x-x1), you replace your pair and get the third option, you have solve the symbols of course for x