Answer:
<h2>A. -2x³ - 6x² + 9x</h2>
Step-by-step explanation:

The answer is 5.5
In a rectangle, lengths of diagonals are equal. If you draw the rectangle, you can see that KM and LN are diagonals, so they must be equal:
KM = LN
KM = 6x + 16
LN = 49
6x + 16 = 49
6x = 49 - 16
6x = 33
x = 33/6
x = 5.5
The answer would be 16 S'mores and the limiting reactant would be the grahams.
(This is assuming that S'mores would need 2 grahams, 1 marshmallow and 1 chocolate piece.)
Limiting reactant would be the reactant that runs out first.
Let's take your problem into account and see what we have:
48 marshallows
32 grahams (2 x 16 per pack)
45 chocolate pieces (5 x 15 pieces per bar)
Since need 2 of the grahams per S'more then the maximum yield of the grahams is 16 S'mores.
The maximum yield of marshmallows is 48.
The maximum yield of chocolate is 45.
Since you cannot make S'mores without the grahams, then you can only make 16 S'mores before the grahams run out.
Answer:
a) The probability that the airline will lose no bags next monday is 0.1108
b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227
c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2
Step-by-step explanation:
The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

a)

The probability that the airline will lose no bags next monday is 0.1108.
b) Note that
. And

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.
c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.
The y axis for dependent variables and x axis for independent