There are three questions related to this problem.
First, the probability of the mail will arrive after 2:30
PM
<span>Find the z-score of 2:30 which is 30 minutes after 2:00.</span>
<span>
z(2:30) = (2:30 – 2:00)/15 = -30/15 = -2</span>
<span>
P(x < 2:30) = P(z<-2) = 0.0228</span>
<span>
</span>
Second, the probability of the mail will arrive at 1:36
PM
<span>Find the z-score of 1:36 which is 24 minutes before 2:00.</span>
<span>
z(1:36) = (1:36 – 2:00)/15 = -24/15 = -1.6</span>
<span>
P(x < 1:36) = P(z<-1.6) = 0.0548</span>
Lastly, the probability of the mail will arrive between 1:48
PM and 2:09 PM
Find the z-score of 1:46 and 2:09 PM which will result to
a z value of 0.034599
<span>P(1:48 < x < 2:09) = P(z<0.034599) = 0.5138</span>
Answers?? need a little bit of a hint of them please.
5/6 can be simplified to 0.83. 10.83*18=194.94 square feet. 1,800 minus 194.94 is 1605.06 more square feet required
We estimate that 12/480=1/40 fish are tagged, so 600 is 1/40 of the total number of fish. This means that there are about 600*40=24000 fish total in the pond.
Answer:
400%
Step-by-step explanation:
Let the sales be "100" in 1994
Since, it decreased 80%, the sales was:
80% = 80/100 = 0.8
0.8 * 100 = 80 (decreased by 80)
So, it was
Sales in 1995: 100 - 80 = 20
Sales in 1996 was same as in 1994, so that's 100
Thus,
Sales in 1994: 100
Sales in 1995: 20
Sales in 1996: 100
We need to find percentage increase form 1995 to 1996, that is what percentage increase is from 20 to 100?
We will use the formula:

Where
New is 100
Old is 20
SO, we have:

So, it increased by 400%