Draw a diagram to illustrate the problem as shown in the figure below.
The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.
The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m
The model for tide depth is

That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours
A plot of the function confirms that the model is correct.
Answer:
Check the explanation
Step-by-step explanation:
Ans=
A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134
M = 10: 1 – 0.9973^10 = 0.0267
M = 20: 1 – 0.9973^20 = 0.0526
M = 30: 1 – 0.9973^30 = 0.0779
M = 50: 1 – 0.9973^50 = 0.126
18)
Ans=
Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values
Answer:
thanks
Step-by-step explanation:
Answer:
-2 and 12
Step-by-step explanation:
To find the two numbers which are 7 units away from 5, add 7 to 5 and subtract 7 from 5
5+7 = 12
5-7 = -2