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jeka94
3 years ago
13

How do you solve interquartile range?

Mathematics
2 answers:
77julia77 [94]3 years ago
5 0
By Hand
Step 1: 
Put the numbers in order.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 2: 
Find the median.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 3: 
Place parentheses around the numbers above and below the median. 
Not necessary statistically, but it makes Q1 and Q3 easier to spot.
(1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27).

Step 4: 
Find Q1 and Q3
Think of Q1 as a median in the lower half of the data and think of Q3 as a median for the upper half of data.
(1, 2, 5, 6, 7),  9, ( 12, 15, 18, 19, 27). Q1 = 5 and Q3 = 18.

Step 5: 
Subtract Q1 from Q3 to find the interquartile range.
18 – 5 = 13.
Kitty [74]3 years ago
3 0
You subtract Quartile 1 and Quartile 3
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  1. A function that models the data is given by this quadratic equation, y = -0.4908x² + 5.8845x + 1.3572.
  2. The number of students that are absent 10 days after the outbreak is equal to 11 students.

<h3>What is a scatter plot?</h3>

A scatter plot can be defined as a type of graph which is used for the graphical representation of the values of two (2) variables, with the resulting points showing any association (correlation) between the data set.

<h3>What is a quadratic function?</h3>

A quadratic function can be defined as a mathematical expression (equation) that can be used to define and represent the relationship that exists between two or more variable on a graph.

In Mathematics, the standard form of a quadratic equation is given by;

ax² + bx + c = 0

By critically observing the graph (see attachment) which models the data in the given table, we can infer and logically deduce that the quadratic function is given by:

y = -0.4908x² + 5.8845x + 1.3572

For the number of students that are absent 10 days after the outbreak, we have:

y = -0.4908(10)² + 5.8845(10) + 1.3572

y = -0.4908(100) + 58.845 + 1.3572

y = -49.08 + 58.845 + 1.3572

Number of students, y = 11.12 ≈ 11 students.

Read more on scatterplot here: brainly.com/question/6592115

#SPJ1

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