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bonufazy [111]
3 years ago
6

Which of the following representations are functions?

Mathematics
1 answer:
allsm [11]3 years ago
4 0

Answer: A and B only

Step-by-step explanation:

In function, for one value of x there must be only one value of y.

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In the triangle shown we can find the angle θ as follows calculator
Xelga [282]
Given a right triangle with hypothenus of measure 34, the side opposite the angle θ of measure 30, and the side adjacent the angle theta of measure 16.

\sin\theta= \frac{opposite}{hypothenuse} = \frac{30}{34} = \frac{15}{17}  \\  \\ \Rightarrow \theta=\sin^{-1}\left( \frac{15}{17} \right) \\  \\  \\ \cos\theta= \frac{adjacent}{hypothenuse} = \frac{16}{34} = \frac{8}{17}  \\  \\ \Rightarrow \theta=\cos^{-1}\left( \frac{8}{17} \right) \\  \\  \\ \tan\theta= \frac{opposite}{adjacent} = \frac{30}{16} = \frac{15}{8}  \\  \\ \Rightarrow \theta=\cos^{-1}\left( \frac{15}{8} \right)
3 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Find the dimensions of a rectangle whose perimeter is 22 meters and whose area is 28 square meters.
UkoKoshka [18]
I hope this helps you

3 0
3 years ago
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The point of a square pyramid is cut off, making each lateral face of the pyramid a trapezoid with the dimensions shown. A trape
aksik [14]
Use the trapezium formula 1/2( sum of parallel lines)( height )

5 0
3 years ago
PLEASE HELPPPP!!!!!!
iragen [17]
The correct answer is A no explanation needed it’s just A
4 0
3 years ago
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