M1q7.) construct a box plot from the data below

There are 16 numbers.

The median is 92.5 ( find the middle two values and divide by 2).

Minumum is 81

Maximum is 109

First quartile is 88.25 (Find median of the lower half of numbers).

Third quartile is 97.75 (Find median of the upper half of numbers.)

The interquartile range is 9.5 ( Difference between the first and third quartile).

Plotting that data in a box plot, the correct one looks like #1

Find the median.

92 and 93 are both middle numbers so add and divide by two.

92 + 93 = 185

185 / 2 = 92.5

Minimum (smallest number): 81

Maximum (largest number): 109

Find the median of the lower values behind the median.

88.25

Find the mean of the higher values ahead of the median.

97.75

Subtract to find the interquartile range.

97.75 - 88.25 = 9.5

The only option with these characteristics is Option A.

Best of Luck!

the slope of the first function is -3. (3/1)

the slope of the second function is 1. (13-10/6-3).

since |-3| is greater than |1|, the function with the steepest slope is the first.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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