Answer: The points are, the top left one, (21,3) and the one on the bottom is (6, 12)
Step-by-step explanation: For the slope, we label one of the points, with y or x. For example, (21, 3) we would label 21 either y1 or x1. Say we put y2, then 3 would be x2. Moving on, (6,12) this time we could label 6 as either y1 or x1. Same thing, but this time we are going to label 6 as y1 and 12 x1. Again, it’s DOES NOT matter what you label them as.
Now to solve it, remember y numbers always go on top! M which is slope. M = y2 - y 1 / x2 - x 1 or with the numbers M = 21 - 6 / 12 - 3 = 15/9 simplified = 5/3
There are 5 couples are sitting in the row
<h3>How many couples are sitting in the row?.</h3>
The given parameters are
Maximum = 6 couples
In the hall, we have the following arrangements
<u>First couple</u>
Right = 2 couples
Left = 1 couple
<u>Another couple</u>
Left = 3 couple
Right = 1 couples
The left of the other couple is the right of the first couple
So, we have the following arrangements:
<u>First couple (1)</u>
Right = 2 couples
Left = 1 couple
<u>Another couple</u>
Right = 1 couples
The total number of couples is
Total = 1 + 2 + 1 + 1
Evaluate the sum
Total = 5
Hence, there are 5 couples are sitting in the row
Read more about combination at:
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Answer:
<h3> Yes, but only if a=0 or b=0</h3>
Step-by-step explanation:
Answer:
a. What is the probability that Carl arrives first?
Probability that Carl arrives first is ¹/₃ = 33.33% since their arrival times is uniformly distributed. The same probability applies to Bob and Alice.
b. What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?
Assuming that Carl arrived on time, 1:10 PM, we must determine the probability that Alice or Bob arrive between 1:20 and 1:30 (half the remaining time)
P = [3 · (¹/₂ - ¹/₃)] · [3 · (¹/₂ - ¹/₃)] = (3 · ¹/₆) · (3 · ¹/₆) = ¹/₂ · ¹/₂ = ¹/₄ = 25% chance that either Alice or Bob arrive more than 10 minutes later
c. What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?
P = 1 - 25% = 75%
d. What is the probability that the person who arrives second will have to wait more than 5 minutes for the third person to show up?
I divided the 20 minutes by 5 to get ¹/₄:
P (|S - T| ≤ ¹/₄) = {[(x + ¹/₄)²] / 2} + (1 / 2x) + {[(⁵/₄ - x)²] / 2} = 0.09375 + 0.25 + 0.09375 = 0.4375 = 43.75%
She ran 1 2/4 more miles, or 1 1/2 or 1.5 miles. I hope this helps!
