6x₁+3x₂=60
6x₁-2x₂=80
Answer:18 outcomes
Step-by-step explanation:
There are 18 outcomes on this scenario card. Students can list all of the outcomes or describe them. Here is the list
(the first number is the outcome from Spinner 1, and the second number is the outcome from Spinner 2).
Answer:
x>13
Step-by-step explanation:
edge 2021
Answer:
1. turkey trolley
2. pumpkin pie
3. family dinner
4. smashed potatoes
5. football
6. wishbones
7. kill the turkey
8. green beans
9. corn you policeman xD
10. black friday
some of the stuff i know but i tried
Step-by-step explanation:
To find the probability that exactly n cards are dealt before the first ace appears, we can use the concept of a geometric distribution. In a geometric distribution, we're interested in the number of trials (in this case, card draws) required for a success to occur (in this case, drawing an ace) for the first time.
The probability of drawing an ace in a single draw from a well-shuffled pack of 52 cards is 4/52 because there are 4 aces out of 52 cards.
So, the probability of drawing a non-ace in a single draw is (52 - 4)/52 = 48/52.
Now, let X be the random variable representing the number of cards drawn before the first ace appears. X follows a geometric distribution with parameter p, where p is the probability of success on a single trial.
P(X = n) = (1 - p)^(n - 1) * p
In this case, p is the probability of drawing an ace on a single trial, which is 4/52, and n is the number of cards drawn before the first ace.
So, the probability that exactly n cards are dealt before the first ace appears is:
P(X = n) = (1 - 4/52)^(n - 1) * (4/52)
Now, to find the probability that exactly k cards are dealt in all before the second ace appears, we need to consider two scenarios:
1. The first ace appears on the nth card, and the second ace appears on the kth card after that. This is represented by P(X = n) * P(X = k).
2. The first ace appears on the kth card, and the second ace appears on the nth card after that. This is represented by P(X = k) * P(X = n).
So, the total probability that exactly k cards are dealt before the second ace appears is:
P(X = n) * P(X = k) + P(X = k) * P(X = n)
You can calculate this probability using the formula for the geometric distribution with p = 4/52 as mentioned earlier for both P(X = n) and P(X = k).