Answer:
75 m, 96 m
Step-by-step explanation:
a. Area = length x width
7050 = 94 x width
7050/94 = w
75 m is width
b. perimeter = 2L + 2W
356 = 2L + 2(82)
356 = 2L + 164
192 = 2L
96 m = L
Assumptions used to construct the model:
-- The coin has two sides.
-- The probability that the coin will land standing on its edge is zero.
-- The probability that the coin will land with one side up or
the other side up is 100% .
-- The coin is "honest". The probability of landing on one side
is equal to the probability of landing on the other side.
Conclusions / Probability model resulting from the assumptions:
The probability of either side up is 1/2 = 0.5 = 50% .
40 tiles
Assuming the vowels and non-vowels are equal to the five picked, there would be 40 tiles that are vowels.
You have picked five different letters. 5% or 1/20 of 100. If two of those five were vowels, then if you consistently pick 2/5 vowels every time, a predicted 40 tiles will be vowels, 60 would not.
If
is the water level after
days, then at the start
we have
. After the first day
, the water level falls by 1.5, so that
. After the second day
, the water level falls by a total of 2(1.5) = 3, so that
. And so on.
The idea is that
has the closed form
(to answer part b).
could be any non-negative number of days, which means we can pick
from the set of non-negative real numbers. However, after a certain point, our function
will start returning negative values, which would translate to the water level falling below the riverbed. So for the purposes of this problem, we should first find out when
reaches 0:
![L(d)=21-1.5d=0\implies d=14](https://tex.z-dn.net/?f=L%28d%29%3D21-1.5d%3D0%5Cimplies%20d%3D14)
That is, the river dries up completely after 14 days. So we can say the domain of
is
.
The range is the set of values that
can take on for the values of
in the domain. In this case, we start at 21 and steadily fall to 0, so the range is
.
After 9 days, the water level falls to
![L(9)=21-9(1.5)=7.5](https://tex.z-dn.net/?f=L%289%29%3D21-9%281.5%29%3D7.5)
so in total, the water level would have subsided by a depth of 21 - 7.5 = 13.5 ft.