B. rolling a 1 on a six-sided number cube
C. getting heads when flipping a coin
D. choosing a X, Y, or Z from a bag containing all letters in the alphabet
To find the coordinates of B, we use the formula for finding the midpoint of a line segment. By plugging in given values and arranging the equation, we get B(8, -3).
To find the coordinates of B, you'd use the formula for the midpoint of a line segment. The midpoint is calculated as follows - ((x1+x2)/2, (y1+y2)/2). Here, the coordinates of the midpoint M(7,-5) and A(6, -7) are given.
Plugging these in the formula:
7=(6 + x2)/2
-5=(-7+ y2)/2
By cross multiplying and arranging the equations, we can find the value of x2 and y2 - these will be the coordinates of point B.
Solving the equations gives B(8, -3) as a result.
#SPJ3
Answer:(8,-3)
Step-by-step explanation:
X2=8 Y2=-3
Answer:
x=-5 i think
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).
A) 4
B) 9
C) 11
D) 20
Answer: 11
Step by step explanation:
Step 1 : solve the first equation to find the value of x
5x + 6 = 10
5x = 10 – 6
5x = 4
X = 4/5
Step 2 : Substitute the value of x in the second equation
10x + 3 = ?
10 ( 4/5) + 3
= 40 / 5 + 3
= 40 + 15 /5
= 55 / 5
= 11