There were 5 of the 15 in the simulation that used a coupon. To find the probability you just divide 5 by 15
P(=>4) = 5/15 = 1/3 - probability of 4 or more
Answer:
Let us say the domain in the first case, has the numbers. And the co-domain has the students, .
Now for a relation to be a function, the input should have exactly one output, which is true in this case because each number is associated (picked up by) with only one student.
The second condition is that no element in the domain should be left without an output. This is taken care by the equal number of students and the cards. 25 cards and 25 students. And they pick exactly one card. So all the cards get picked.
Note that this function is one-one and onto in the sense that each input has different outputs and no element in the co domain is left without an image in the domain. Since this is an one-one onto function inverse should exist. If the inverse exists, then the domain and co domain can be interchanged. i.e., Students become the domain and the cards co-domain, exactly like Mario claimed. So, both are correct!
I think The last one in right
y |2| 6 |10 |14
x |6 |18 |30 |42
1/5 of 200 is 40 which mean she gave 40 away and she has a 160 left
We need to evaluate the value of fraction 
.
We also given j=12.
In order to evaluate the value of fraction , we need to plug j=12.
Plugging j=12, we get
We have 12 in numerator and 4 in denominator.
We always divide top number by bottom number.
So, we need to divide 12 by 4.
On dividing 12 by 4 we get 3.
<h3>Therefore,
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