Answer:
The probability is 0.3576
Step-by-step explanation:
The probability for the ball to fall into the green ball in one roll is 2/1919+2 = 2/40 = 1/20. The probability for the ball to roll into other color is, therefore, 19/20.
For 25 rolls, the probability for the ball to never fall into the green color is obteined by powering 19/20 25 times, hence it is 19/20^25 = 0.2773
To obtain the probability of the ball to fall once into the green color, we need to multiply 1/20 by 19/20 powered 24 times, and then multiply by 25 (this corresponds on the total possible positions for the green roll). The result is 1/20* (19/20)^24 *25 = 0.3649
The exercise is asking us the probability for the ball to fall into the green color at least twice. We can calculate it by substracting from 1 the probability of the complementary event: the event in which the ball falls only once or 0 times. That probability is obtained from summing the disjoint events: the probability for the ball falling once and the probability of the ball never falling. We alredy computed those probabilities.
As a result. The probability that the ball falls into the green slot at least twice is 1- 0.2773-0.3629 = 0.3576
Answer:
8
Step-by-step explanation:
if you substitute 3 for x the equation becomes y=2(3) + 2
using order of operations you multiply 3 by 2 and then add 2 which will give you 8
What is the question im solving x for?
Answer:
The probability is 1/2
Step-by-step explanation:
The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.
When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.
As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words
The values of the cummulative distribution of the Standard Normal distribution, lets denote it 
, are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization
Answer:
$27.00
Step-by-step explanation:
First, 10% of 30 = 3.
Second, 30 - 3 = 27
Hope this helps!
