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tresset_1 [31]
3 years ago
15

What is the median of the following data set?

Mathematics
2 answers:
AysviL [449]3 years ago
8 0

Answer:

0 is the answer

Step-by-step explanation:

natali 33 [55]3 years ago
5 0
0 is the answer my friend.
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HELP ASAP 6TH GRADE MATH I WILL GIVE BRAINLIEST
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It would be letter a
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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
5x2 – 15x – 20. factorise
hoa [83]

Answer:

5(x−1)(x+4)? i think

Step-by-step explanation:

7 0
3 years ago
(−2k <br> 3<br> −7k <br> 2<br> +5k)+(6k <br> 2<br> +3k)=
shusha [124]

Answer: 5k +7

Step-by-step explanation: -2k + -7k + 5k is -4K and 2+3=5 -4K +5 and 6k +3k= 9k + 2 then u do -4K plus 9k and get 5k next, 5+2

7 0
2 years ago
A.A fish is 12 meters below the surface of the ocean. What is its elevation?
dalvyx [7]
A. -12
b. +28
c. 40meters
5 0
2 years ago
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