Answer:
I believe it is 0.5
Step-by-step explanation:
If you flip a normal coin (called a “fair” coin in probability parlance), you normally have no way to predict whether it will come up heads or tails. Both outcomes are equally likely. There is one bit of uncertainty; the probability of a head, written p(h), is 0.5 and the probability of a tail (p(t)) is 0.5. The sum of the probabilities of all the possible outcomes adds up to 1.0, the number of bits of uncertainty we had about the outcome before the flip. Since exactly one of the four outcomes has to happen, the sum of the probabilities for the four possibilities has to be 1.0. To relate this to information theory, this is like saying there is one bit of uncertainty about which of the four outcomes will happen before each pair of coin flips. And since each combination is equally likely, the probability of each outcome is 1/4 = 0.25. Assuming the coin is fair (has the same probability of heads and tails), the chance of guessing correctly is 50%, so you'd expect half the guesses to be correct and half to be wrong. So, if we ask the subject to guess heads or tails for each of 100 coin flips, we'd expect about 50 of the guesses to be correct. Suppose a new subject walks into the lab and manages to guess heads or tails correctly for 60 out of 100 tosses. Evidence of precognition, or perhaps the subject's possessing a telekinetic power which causes the coin to land with the guessed face up? Well,…no. In all likelihood, we've observed nothing more than good luck. The probability of 60 correct guesses out of 100 is about 2.8%, which means that if we do a large number of experiments flipping 100 coins, about every 35 experiments we can expect a score of 60 or better, purely due to chance.
3/4 = .75 and 1 1/2 = 1.5
The least number on that line is -1 and the greatest is 1.5
The order is -1, -.9, 3/4, 1.1, and 1 1/2
Answer:
5x-12 > with the _ on the bottom 6+2x
Step-by-step explanation:
x is the number/variable
Answer:
31/40
Step-by-step explanation:
The question is incomplete. Here is the complete question with appropriate diagram.
The circle below has an area of 314 square centimeters, and the square inside the circle has a side length of 2 centimeters.
What is the probability that a point chosen at random is in the blue region?
Given the area of the circle to be 314cm², we need to get the diameter of the circle first since the diameter of the circle is equivalent to length of the side of the square inscribed in it.
Using the formula Area of a circle = πr²
314 = 3.14r²
r² = 314/3.14
r² = 100
r = 10 cm
Diameter of the circle = 2*10 = 20 cm
Area of a square = Length * length
Area of the outer square = 20*20 = 400cm²
Area of the inner square with side length 2cm = 2*2 =4cm²
Area of the shaded region = Area of the square - Area of the inner square
= 314-4 = 310cm²
The probability that a point chosen at random is in the blue region = Area of the shaded region/total area of the outer square
= 310/400
= 31/40
x=3 and y=2 because 6+6 = 12 and 2 times 3 is 6 :)
