Answer: the answer is “B”
Step-by-step explanation: answer is LETTER B
This problem is a combination of the Poisson distribution and binomial distribution.
First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961
For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181
3minutes 45 seconds = 3 45/60 = 3.75 minutes
28 copies/ 1 minute = x copies/3.75 minutes
using cross products
28 * 3.75 = 1 * x
105 = x
We can print 105 copies in 3 3/4 minutes
Answer:
“Oh, the bells, bells, bells!
What a tale their terror tells
Of Despair!
How they clang, and clash, and roar!
What a horror they outpour
On the bosom of the palpitating air!
Yet the ear it fully knows,
By the twanging,
And the clanging,
How the danger ebbs and flows;
Yet the ear distinctly tells,
In the jangling,
And the wrangling.”
Step-by-step explanation:
Answer:
≈50.6
Step-by-step explanation:
Not sure what precision level this problem is looking for, but for right-skewed distributions, we know that the mean is going to be pulled right and therefore the mean should be larger than the median. To a high confidence level, the mean should fall between 50 and 59, or in the third column.
If a single estimation is wanted, assume the values inside each column are uniformly distributed:
