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boyakko [2]
3 years ago
7

Which expression below gives the average rate of change of the function g(x) = -x2 - 4x on the interval 6 ≤ x ≤ 8 ?

Mathematics
2 answers:
Scrat [10]3 years ago
7 0

Answer:

the first selection (see below)

Step-by-step explanation:

The average rate of change (m) on the interval [x1, x2] is given by ...

... m = (g(x2) -g(x1))/(x2 -x1)

For g(x) = -x²-4x and (x1, x2) = (6, 8), the expression is the one attached.

Papessa [141]3 years ago
3 0

Answer:

\frac{[-8^2 - 4(8)]- [-6^2 - 4(6)]}{8-6}

Step-by-step explanation:

average rate of change of the function g(x) = -x^2 - 4x on the interval 6 ≤ x ≤ 8

To find average rate of change we use formula

Average =\frac{g(x_2)-g(x_1)}{x^2-x_1}

use the given interval 6<=-x<=8

x2=8  and x1= 6

we replace the value in the given formula

g(x) = -x^2 - 4x

g(8) = -8^2 - 4(8)

g(6) = -6^2 - 4(6)

x2-x1 is 8-6

So equation becomes

\frac{[-8^2 - 4(8)]- [-6^2 - 4(6)]}{8-6}

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A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part
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Answer:

The standard deviation increased but there was no change in the interquantile range          

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{8726}{20} = 436.3

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

S.D = \sqrt{\frac{75924.2}{19}} = 63.21

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

Mean =\displaystyle\frac{8406}{20} = 420.3

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

S.D = \sqrt{\frac{247636.2}{19}} = 114.16

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

Thus. the standard deviation increased but there was no change in the interquantile range.

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