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SOVA2 [1]
3 years ago
5

Average mido Find the mean, median, and range of the following data items: 74, 75, 78, 84,94 18 20 84 78 81 82.5 Mean Median Ran

ge​

Mathematics
1 answer:
Zarrin [17]3 years ago
3 0

Answer:

18: mean = 70.5

median = 76.5

range = 76

20:

mean = 70.83

median = 76.5

range = 76

84:

mean = 81.5

median = 79.5

range = 20

78:

mean = 80.5

median = 81

range = 20

81:

mean = 81

median = 80.25

range = 20

82.5:

mean = 81.5

median = 80.25

range = 20

Step-by-step explanation:

mean = average = sum of numbers / amount of numbers present

median = middle number (arranged least to greatest)

range = largest number - smallest number

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Help me please i put numbers next to the answers you make it easier to answer insted of typing it:)
Brut [27]
AB = 4/3
BC = -2/7
CD = 4/3
AD = -2/7

Parallelogram because both sides are parallel
7 0
3 years ago
- 8x + 64 = 56-X<br> what’s the answer
Readme [11.4K]

Step-by-step explanation:

two answers either

x = 8

or

 x = -1

6 0
3 years ago
Read 2 more answers
Solve 2 log(x) + log(5) = log (80) using the one-to-one property.
vodomira [7]

x=4

Step-by-step explanation:

2log(x)+log(5)=log(80)

log(x^2)+log(5)=log(80)

log(5x^2)=log(80)

5x^2 =80

x^2 =80/5

x^2=16 , taking square root both sides

x=√16

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5 0
3 years ago
The caller times at a customer service center has an exponential distribution with an average of 22 seconds. Find the probabilit
jenyasd209 [6]

Answer:

The probability that a randomly selected call time will be less than 30 seconds is 0.7443.

Step-by-step explanation:

We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.

Let X = caller times at a customer service center

The probability distribution (pdf) of the exponential distribution is given by;

f(x) = \lambda e^{-\lambda x} ; x > 0

Here, \lambda = exponential parameter

Now, the mean of the exponential distribution is given by;

Mean =  \frac{1}{\lambda}  

So,  22=\frac{1}{\lambda}  ⇒ \lambda=\frac{1}{22}

SO, X ~ Exp(\lambda=\frac{1}{22})  

To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;

    P(X\leq x) = 1 - e^{-\lambda x}  ; x > 0

Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)

        P(X < 30)  =  1 - e^{-\frac{1}{22} \times 30}

                         =  1 - 0.2557

                         =  0.7443

7 0
4 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
3 years ago
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