Answer:
The probability that a randomly selected adult is either overweight or obese is 0,688
Step-by-step explanation:
Probability that an american adult is overweight = 0,331
Probability that an american adult is obese = 0,357
Let's find the probability that an adult is either overweight or obese, that means both events are mutually exclusive. We are interested in the probability that just one event occurs, that probability is the sum of their individual probabilities
P(overweight ∪ obese) = P(overweight) + P(obese) = 0,331 + 0,357 = 0,688
The statement that must be true for the given condition is <span>m∠X + m∠Y < 90°</span>
The <u>different</u> selections of<u> one</u> meat and <u>one</u> vegetable are possible are 18 selections.
Since the Hickory Stick has a selection of 3 meats and 6 vegetables, the number of ways we can select <u>one </u>meat out of 3 is ³C₁ = 3.
Also, the number of ways we can select <u>one </u>vegetable out of 6 is ⁶C₁ = 6.
So, the total number of selections of <u>one</u> meat and <u>one</u> vegetable is ³C₁ × ⁶C₁ = 3 × 6
= 18 selections
So, the <u>different</u> selections of<u> one</u> meat and <u>one</u> vegetable are possible are 18 selections.
Learn more about combination here:
brainly.com/question/19341024
Answer:
17 + 23 = 40
23 - 17 = 6 the other person did it right.
Answer:
x=4
Step-by-step explanation:
Consider the coordinate pair (4,y)
We observe that,
y can take any real values to lie on the graph of the equation.
As y really isn't a constant, we exclude it and take the equation as x=4
