0.07 is what of 0.7
1 answer:
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Answer:
S(t) = a.sin (b.t) + d
a = -1.5, b = (2π/365), d = 3.47
S(t) = -1.5 sin (2πt/365) + 3.47
Step-by-step explanation:
Complete Question is presented in the attached image to this solution.
- Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function S(t) of time t (in days) using a sinusoidal expression of the form
S(t) = a.sin(b.t) + d.
On day t = 0, the stock is at its average value of $3.47 per share, but 91.25 days later, its value is down to its minimum of $1.97.
Find S(t). t should be in radians.
S(t) =
Solution
S(t) = a.sin(b.t) + d.
At t = 0, S(t) = $3.47
S(0) = a.sin(b×0) + d = a.sin 0 + d = 3.47
Sin 0 = 0,
S(t=0) = d = 3.47.
At t = 91.25 days, S(t) = $1.97
But, it is given that T has to be in radians, for t to be in radians, the constant b has to convert t in days to radians.
Hence, b = (2π/365)
S(91.25) = 1.97 = a.sin(b×91.25) + d
d = 3.47 from the first expression
S(t = 91.25) = a.sin (91.25b) + 3.47 = 1.97
1.97 = a.sin (2π×91.25/365) + 3.47
1.97 = a sin (0.5π) + 3.47
Sin 0.5π = 1
1.97 = a + 3.47
a = -1.5
Hence,
S(t) = a.sin (b.t) + d
a = -1.5, b = (2π/365), d = 3.47
S(t) = -1.5 sin (2πt/365) + 3.47
Hope this Helps!!!
Answer:
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618
Step-by-step explanation:
Normal with mean 6.88 minutes and
a standard deviation of 0.57 minutes.
Choose a student at random from this group and call his time for the mile Y. Find P(Y<6)
y ≈ normal (μ, σ)
The z score is the value decreased by the mean divided by the standard deviation
Therefore the probability that a randomly selected student has time for mile run is less than 6 minute is 0.0618