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bogdanovich [222]
3 years ago
14

Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 132,127,106,140,158,135,129,138 med

ian: first quartile: third quartile: interquartile range
Mathematics
1 answer:
Ket [755]3 years ago
4 0

Answer:

133.5, 127.5, 139.5, 12

Step-by-step explanation:

order data:

106, 127, 129, 132, 135, 138, 140, 158

Median:

The middle number: (8+1)/2 = 4.5 between the 4&5 numbers

= (135-132)/2= 1.5

1.5 + 132 = 133.5

lower quartile (1st quartile):

(8+1)/4 = 2.25 between the 2&3 numbers

(129+127)/4=0.5

0.5+127 = 127.5

upper quartile(3rd quartile):

(8+1)/4 x3 = 6.75 between the 6&7 numbers

(140-138)/4 x3 = 1.5

1.5 + 138 =139.5

Interquartile range:

139.5-127.5= 12

hope this helps

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The demand equation illustrates the price of an item and how it relates to the demand of the item.

  • The slope of the demand function is -1/2
  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

From the question, we have:

Plates = 300

Price = 20

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}

The slope (m) is calculated using:

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The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

R(x) = (300 - 10x) \times (20 + 5x)

Open the brackets to calculate the maximum revenue

R(x) =6000 + 1500x - 200x - 50x^2

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Divide by 100

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  • The equation of the demand function is: R(x) = (300 - 10x) \times (20 + 5x)
  • The price that maximizes her revenue is: Ghc 85

Read more about demand equations at:

brainly.com/question/21586143

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