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3 years ago

Identify the frequency of the graph

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

ANSWER

$\frac{1}{2}$

EXPLANATION

The frequency of a wave function refers to the number of times the graph repeats its cycle within the interval [0,2π]

We can observe explicitly from the graph that the frequency is half.

The period of the given function is 4π

$\frac{2\pi}{b} = 4\pi$

$b = \frac{2\pi}{4\pi} = \frac{1}{2}$

Hence the frequency is half.

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Read 2 more answers

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

exis [7]

Answer:

(A) The probability that the number of drivers will be at most 10 is 0.0108.

(B) The probability that the number of drivers will exceed 20 is 0.4409.

(C) The probability that the number of drivers will be between 10 and 20 (inclusive) is 0.5483 and excluding 10 and 20 is 0.4595.

(D) The probability that the number of drivers will within 2 standard deviations of the mean is 0.9549.

Step-by-step explanation:

Let X = number of drivers who travel a distance during a designated time period.

The random variable X follows a Poisson distribution with parameter λ = 20.

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

(A)

Compute the probability that the number of drivers will be at most 10 as follows:

P (X ≤ 10) = P (X = 0) + P (X = 1) + P (X = 2) ... + P (X = 10)

$=\frac{e^{-20}20^{0}}{0!}+\frac{e^{-20}20^{1}}{1!}+...+\frac{e^{-20}20^{9}}{9!}+\frac{e^{-20}20^{10}}{10!}\\=0.00+0.00+...+0.00291+0.00582\\=0.0108$

Thus, the probability that the number of drivers will be at most 10 is 0.0108.

(B)

Compute the probability that the number of drivers will exceed 20 as follows:

P (X > 20) = 1 - P (X ≤ 20)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) ... + P (X = 19) + P (X = 20)]

$=1-[\frac{e^{-20}20^{0}}{0!}+\frac{e^{-20}20^{1}}{1!}+...+\frac{e^{-20}20^{19}}{19!}+\frac{e^{-20}20^{20}}{20!}]\\=1-[0.00+0.00+...+0.0884+0.08884]\\=0.4409$

Thus, the probability that the number of drivers will exceed 20 is 0.4409.

(C)

Compute the probability that the number of drivers will be between 10 and 20 (inclusive) as follows:

P (10 ≤ X ≤ 20) = P (X ≤ 20) - P (X ≤ 10)

$=0.5591-0.0108\\=0.5483$

Compute the probability that the number of drivers will be exactly between 10 and 20 as follows:

P (10 < X < 20) = P (X = 11) + P (X = 12) + ... + P (X = 19)

$=\frac{e^{-20}20^{11}}{11!}+\frac{e^{-20}20^{12}}{12!}+\frac{e^{-20}20^{13}}{13!}+...+\frac{e^{-20}20^{19}}{19!}\\=0.01058+0.01763+0.02712+...+0.08884\\=0.4595$