Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of
2 answers:
Answer:
Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)
Step-by-step explanation:
Given: Probability of a children under age 5 has food allergies = 7%
=
To find : Probability of atleast one of 12 student has food allergies
Probability of a chindren under age 5 does not have food allergies =
⇒ prob =
now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.
But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)
Probability of 0 student having allergy out of 12 student =
Therefore, Probability of atleast one of 12 student has food allergies
=
=
≈ 0.58 ( approx)
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Answer:
The measure of angle JKL is 97 degrees.
Step-by-step explanation:
Given:
A triangle JKL .
Measure of exterior angle K = 83 degrees
We have to find the measure of .
In we see that the exterior angle is in linear pair with
.
Note:
Linear pair angles sum is 180 degree.
Let be 'x' degrees .
So,
⇒
⇒ <em>Subtracting 83 from both sides.</em>
⇒
⇒
⇒ degrees
The measure of angle JKL in the triangle is of 97 degrees.
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is . Thus, there are only 35 ways to distribute the blackboards in this case.