The answer is linked below
Answer:
1/3,11/12,9/8
Step-by-step explanation:
Hi! Let's start off with creating common denominators for all three of these fractions.
The LCM (least common multiple) of these three denominators is 24, which means that the three numbers you see at the bottom of the fraction (12, 8, and 3) all in some way have a number that will cause it to multiply to 24, and will make it much more easier to determine the fraction.
1) Change all of the denominators into 24 by multiplying the right number
11/12×2/2=>22/24
9/8×3/3=>27/24
1/3×8/8=>8/24
So now I want you to imagine that the denominators don't exist. From there, you can sort out the fractions from least to greatest! Hope you understood this :)
The numbers that are irrational are ∛4 and 4 + √5
<h3>How to determine which of the numbers are irrational?</h3>
The numbers are given as:
negative four and 8237 ten thousandths, one half pi, cube root of four, and four plus the square root of twenty-five
Rewrite properly as:
-4.8237, π/2, ∛4 and 4 + √5
When the above numbers are evaluated, we have:
-4.8237 = -4.8237
π/2 = 11/7
∛4 = 1.5874.....
4 + √5 = 6.23606.....
Irrational numbers are numbers that cannot be represented as a fraction of two integers and they are always non-terminating decimals
∛4 and 4 + √5 are non-terminating decimals
Hence, the numbers that are irrational are ∛4 and 4 + √5
Read more about irrational numbers at:
brainly.com/question/8798082
#SPJ1
Answer:
a) No
b) 42%
c) 8%
d) X 0 1 2
P(X) 42% 50% 8%
e) 0.62
Step-by-step explanation:
a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.
b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6
P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7
P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
c) P(win first game) = 0.4
P(win second game) = 0.2
P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%
d) X 0 1 2
P(X) 42% 50% 8%
P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%
e) The expected value 
f) Variance 
Standard deviation 
