Answer:
do you have an attachment or anything?
Step-by-step explanation:
Answer:
The missing reason is Subtraction Property of Equality.
Step-by-step explanation:
⇒ (Given)
On Solving we get;
Multiplying 2 on both side we get;
⇒ (Multiplication Property of Equality)
Now Subtracting Both side by 10 we get;
⇒ (Subtraction Property of Equality)
Now Dividing both side by 3 we get;
⇒ (Division Property of Equality)
Hence the missing reason is Subtraction Property of Equality.
A = event the person got the class they wanted
B = event the person is on the honor roll
P(A) = (number who got the class they wanted)/(number total)
P(A) = 379/500
P(A) = 0.758
There's a 75.8% chance someone will get the class they want
Let's see if being on the honor roll changes the probability we just found
So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).
---------------
A and B = someone got the class they want and they're on the honor roll
P(A and B) = 64/500
P(A and B) = 0.128
P(B) = 144/500
P(B) = 0.288
P(A | B) = P(A and B)/P(B)
P(A | B) = 0.128/0.288
P(A | B) = 0.44 approximately
This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.
Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.
Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.
Working attached below:
Hope this helps! Any questions let me know :)
Answer:
1/4
Step-by-step explanation:
The chance of it landing heads up once would be 1/2
The chance of it landing heads up twice would be 1/4
The chance of it landing heads up three times would be 1/8
The chance of it landing heads up four times would be 1/16
See the pattern?
Pattern: <em><u>Divide the denominator by 2.</u></em>
<h2>
<em>Please give brainliest!!! :) </em></h2>
