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2 years ago

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If k(x) = 5x – 6, which expression is equivalent to (k + k)(4)?. . 5(4 + 4) – 6. 5(5(4) – 6) – 6. 54 – 6 + 54 – 6. 5(4) – 6 + 5(

4) – 6

2 answers:

Dmitrij [34]2 years ago

6 0

K(x) = 5x-6
(k+k)(x) = k(x) + k(x) = 5x-6 +5x-6
just plug in x = 4,
(k+k)(4) = 5(4) -6 + 5(4) -6
thats your answer, the last option.

Ghella [55]2 years ago

5 0

Answer:

d

Step-by-step explanation:

100%

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Answer:

a) $\bar X_A =22.744$

$\bar X_B =20.7$

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$\bar X_D =18.306$

b) $Median_A =\frac{23+24}{2}=23.5$

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c) $\bar X_A =23.25$

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$\bar X_D =20.69$

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

If we apply this formula for the four methods we got:

$\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

$Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

$\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

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