Answer:
The probability that the page will get at least one hit during any given minute is 0.9093.
Step-by-step explanation:
Let <em>X</em> = number of hits a web page receives per minute.
The random variable <em>X</em> follows a Poisson distribution with parameter,
<em>λ</em> = 2.4.
The probability function of a Poisson distribution is:
Compute the probability that the page will get at least one hit during any given minute as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
Thus, the probability that the page will get at least one hit during any given minute is 0.9093.
Triangle ABD and ADC are righttriangles so u can use the Pythagorean theorem/triples. the triples are 3/4/5 and 5/12/13. 12 is 4x3, and 15 is 5x3, so BD is 3x3, which is 9. AC is 13, as it is part of the Pythagorean’s triples
Is it subtract 228? if so then 3-5 equals -2. -2-8 equals -10. -10-228 equals -238
It is rotated 90 degress clockwise then reflected across the x-axis
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
& 
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
= 
= 
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4
= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 - 
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
