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
983.625
9 = hundreds
8 = tens
3 = ones
6 = tenths
2 = hundredths
5 = thousandths
You want to round to the nearest hundredths
Look at the number on the right, if it's bigger than 5, round up, if it's 4 or lower, round down.
In this case, the number to the right is a 5, so you round the 2 (hundredths) up one so:
<span>983.63</span>
Answer:
g(6)=-16
Step-by-step explanation:
g(6)=-2(6)-4
g(6)=-12-4
g(6)=-16
Hope this helps lol
The easiest way to tell whether lines are parallel, perpendicular, or neither is when they are written in slope-intercept form or y = mx + b. We will begin by putting both of our equations into this format.
The first equation,

is already in slope intercept form. The slope is 1/2 and the y-intercept is -1.
The second equation requires rearranging.

From this equation, we can see that the slope is -1/2 and the y-intercept is -3.
When lines are parallel, they have the same slope. This is not the case with these lines because one has slope of 1/2 and the other has slope of -1/2. Since these are not the same our lines are not parallel.
When lines are perpendicular, the slope of one is the negative reciprocal of the other. That is, if one had slope 2, the other would have slope -1/2. This also is not the case in this problem.
Thus, we conclude that the lines are neither parallel nor perpendicular.
Answer:
the numbers are going down by .5 nvm I answered this wrong.