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Answer:
The box-and-whisker plot of this distribution is presented in the attached image to this solution.
Step-by-step explanation:
A box plot gives a visual representation of the distribution of the data, showing where most values lie and those values that greatly differ from the rest, called outliers.
A box and whiskers plot shows 5 major information about the distribution of data. It shows:
- The maximum variable.
- The minimum variable.
- The Median.
- The first quartile.
- The third quartile.
Further info such as the range and Inter quartile range can then be obtained from this 5-number summary.
The elements of the box plot are described thus;
The bottom side of the box represents the first quartile, and the top side, the third quartile. Therefore, the width of the central box represents the inter-quartile range.
The horizontal line inside the box is the median.
The lines extending from the box reach out to the minimum and the maximum values in the data set, as long as these values are not outliers. The ends of the whiskers are marked by two shorter horizontal lines.
Variables in the dataset, higher than Q3+(1.5×IQR) or lower than Q1-(1.5×IQR) are considered outliers and are usually shown using dots above the top whisker or below the bottom whisker.
The required boxplot for this question is given in the attached image to this solution.
The median for the boxplot isn't provided, but it was assumed to be midway between the first and third quartile.
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By applying the segment addition postulate, the <u>value of v = 7</u>
- According to the Segment Addition Postulate, it holds that if point C is between points D and E, therefore:
DC + CE = DE
- Given that:
- Therefore, by substitution, we will have the following equation:
- Open the bracket and solve for the value of v.
- Add like terms
- Add 23 to both sides
- Divide both sides by 5
v = 10
Therefore, using the segment addition postulate, the <u>value of v = 10</u>
Learn more about the segment addition postulate here:
Answer:
Isosceles triangle
Step-by-step explanation:
It is assumed that the triangle is a right angle triangle, implying that Pythagoras theorem becomes applicable.
If the first side is represented by x and the second side by y,
x + y = 5
The hypotenuse is always the longest side hence it can be presumed to be 4 since both x and y cannot be more than 5.
Therefore using pythagoras equation:
hyp^2 = sqrt(opp^2 + adj^2)
this implies that
therefore
16 = sqrt (x^2 + y^2) ----------------- (i)
since x + y = 5, y = 5 - x ------------------- (ii)
substituting (ii) into eqn (i)
16 = sqrt ( x^2 + (5-X)^2)
16 = sqrt( x^2 + 25 -10x + x^2)
16 = sqrt( x^2 + x^2 - 10x + 25)
16 = sqrt( 2x^2 - 10x + 25)
squaring both sides
256 = 2x^2 - 10x + 25
this can be refined as
2x^2 - 10x - 231 = 0
Solving using quadratic equation:
where a = 2, b = -10 and c = -231
x = 13.53 or -8.53.
Since x is a side of a triangle that cannot be negative,
x = 13.53 but x can never be greater than 5 since x + y = 5
This implies that the triangle is not a right angle triangle; it is most likely an isosceles triangle with two sides of equal length such that x = y = 2.5 inches while the longest side; the hypotenuse is 4 inches.
substituting for x in