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.
For independent events,
P(X ∩ Y) = P(X)* P(Y)
=>
1/3 = P(X)*(5/6)
solve for P(X) =>
P(X) = (1/3)*(6/5) = 2/5 = 0.4
Answer:
Option B.
Step-by-step explanation:
Given information:
Σ(x − M) = 44
where, M is mean.
Sample size = 12
The computational formula for sample variance is

where, M is sample mean and N is sample size.
Substitute Σ(x − M) = 44 and N=12 in the above formula.



The sample variance is 4.0.
Therefore, the correct option is B.
Answer:
The quantity of water in the tank after 15 days is 1610.0 gallons OR 1.61 × 10³ gallons.
Step-by-step explanation:
The amount of water in the tank after 15 days is given by the series
910+(−710)+810+(−610)+⋯+310+(−110)+210
From the series, we can observe that, if water is added for a particular day then water will be drained the following day.
Also, for a day when water is to be added, the quantity to be added will be 100 gallon lesser than the quantity that was last added. Likewise, for a day when water is to be drained, the quantity to be drained will be 100 gallons lesser than the quantity that was last drained.
Hence, we can complete the series thus:
910+(−710)+810+(−610)+710(-510)+610(-410)+510(-310)+410(-210)+310+(−110)+210
To evaluate this, we get
910-710+810-610+710-510+610-410+510-310+410-210+310-110+210
= 1610.0 gallons
Hence, the quantity of water in the tank after 15 days is 1610 gallons OR 1.61 × 10³ gallons.
Answer
B) x + 20 ≥ 50
Step-by-step explanation: