What is the difference between the mean and median (rounded to the nearest tenth) of these numbers 6,6,6,7,7,7,8,8,8,9

1 answer:

Nadusha1986 [10]3 years ago

3 0

The median is the moddle number of a data set. If the number of numbers is even add the 2 middle numbers and divide by 2.

Median = (7+7)÷2 = 7

The mean is the averagw value in a data set. You find it be adding all the numbers and then divide the answer by the number of numbers.

Mean: (6+6+6+7+7+7+8+8+8+9) ÷ 10

= 72 ÷ 10 = 7.2

Happy to help! Marking my answer as the Brainliest would really be aprreciated.

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Help more than 10 points

Drupady [299]

Ratio triangle height to square side 7:8
ratio triangle base to square side 1:2
Area of the square is 64 in² then the sides = 8 in

ratios give
h = 7(8)/8
h = 7

b = 8/2
b = 4

Area Triangle = 0.5(7)(4)
= 14

Then shaded region is the Area of the square minus the Area of the triangle.
A = 64 - 14
A = 50 in²

3 0

4 years ago

Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$