Answer:
3
Step-by-step explanation:
The computation of the mean is shown below:
= Total sum ÷ number of observations
= (1 + 2 + 2 + 2 + 3 + 1 + 4 + 0 + 5 + 3 + 6 + 2) ÷ 12
= 31 ÷ 12
= 2.58
= 2.6
= 3
Answer:
(2.4, -1.2)
Step-by-step explanation:
Start by moving the x and the y to the same side and moving the number across the equal sign in both equations. We should now have y-0.45x=-2.3 and 2y+4.2x=7.8. We can use the elimination method by multiplying the first equation by -2 to get -2y+0.9x=4.6 and 2y+4.2x=7.8. From there, add the two equations together, eliminating y (-2+2=0). We now have 5.1x=12.4; divide both sides by 5.1 to get x=2.4. Then, in any of the two equations, let's use y-0.45x=-2.3, substitute x with 2.4. Now we have y-1.08=-2.3. Add 1.08 to both sides to get y=-1.22; round that to the nearest tenth to get -1.2.
Answer:
The binomial probability formula can not be used for this experiment because it does not state the number of times he expects to draw his favorite suit.
Step-by-step explanation:
The binomial probability formula is expressed as follows:
P (k success in n trials) =


n = number of trials, k = number of successes, n-k = number of failures, p = probability of success in one trial and q = 1 - p = probability of failure in one trial.
In the given problem, all of the variables are known except for 'k', the amount of times that the student predicts he will draw his favorite suit.
Answer: D.
Step-by-step explanation: If you look closely, putting a 100 wouldn't make sense
Answer:
0.8413 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.00 centimeters
Standard Deviation, σ = 0.60 centimeters
Sample size, n = 16
We are given that the distribution of average diameter is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =

P(diameter of sample is more than 3.85 centimeter)
P(x > 3.85)
Calculation the value from standard normal z table, we have,
0.8413 is the probability that the the average diameter of sample of 16 sand dollars is more than 3.85 centimeters.