Option a is correct. The calculated answer is 0.150
<h3>How to get the value using the cdf</h3>
In order to get P(0.5 ≤ X ≤ 1.5).
This can be rewritten as
p = 0.5
and P = 1.5
We have the equation as
This would be written as
1.5²/16 - 0.5²/16
= 0.1406 - 0.015625
= 0.124975
This is approximately 0.1250
Read more on cdf here:
brainly.com/question/19884447
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<h3>complete question</h3>
Use the cdf to determine P(0.5 ≤ X ≤ 1.5).
a) 0.1250
b) 0.0339
c) 0.1406
d) 0.0677
e) 0.8750
f) None of the above
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
Answer:
the answer is that both x and y are acute
<h2>
Answer:</h2><h2>40 minutes</h2>
Step-by-step explanation:
