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3 years ago

What are the more appropriate measures of center and spread for this data set

Mathematics

2 answers:

MArishka [77]3 years ago

7 0

Answer:

(B) and (D)

Step-by-step explanation:

A box plot or boxplot is a method for graphically depicting groups of numerical data through their quartiles.

Also, we know that in a box plot, the center is measured in the form of median and the spread can be measured in the form of Interquartile range which can be calculated by calculating the value of the upper and lower quartiles.

Thus, the best measure to find the center of the data represented in box plot is median and the best measure to find the spread is Interquartile range.

Hence, options B and D are correct.

Leya [2.2K]3 years ago

5 0

Answer:

I would go with answer C and A

Hope this helped :)

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Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters a

Agata [3.3K]

Answer:

The answers to the question is

(a) Jamie is gaining altitude at 1.676 m/s

(b) Jamie rising most rapidly at t = 15 s

At a rate of 2.094 m/s.

Step-by-step explanation:

(a) The time to make one complete revolution = period T = 15 seconds

Here will be required to develop the periodic motion equation thus

One complete revolution = 2π,

therefore the we have T = 2π/k = 15

Therefore k = 2π/15

The diameter = radius of the wheel = (diameter of wheel)/2 = 5

also we note that the center of the wheel is 6 m above ground

We write our equation in the form

y = $5*sin(\frac{2*\pi*t}{15} )+6$

When Jamie is 9 meters above the ground and rising we have

9 = $5*sin(\frac{2*\pi*t}{15} )+6$ or 3/5 = $sin(\frac{2*\pi*t}{15} )$ = 0.6

which gives sin⁻¹(0.6) = 0.643 = $\frac{2*\pi*t}{15}$

from where t = 1.536 s

Therefore Jamie is gaining altitude at

$\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =$ 1.676 m/s.

(b) Jamie is rising most rapidly when the velocity curve is at the highest point, that is where the slope is zero

Therefore we differentiate the equation for the velocity again to get

$\frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15})$ =0, π, 2π

Therefore $-sin(\frac{2\pi t}{15} )$ = 0 whereby t = 0 or

$\frac{2\pi t}{15}$ = π and t = 7.5 s, at 2·π t = 15 s

Plugging the value of t into the velocity equation we have

$\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) =$ - 2/3π m/s which is decreasing

so we try at t = 15 s and we have $\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pi$ m/s

Hence Jamie is rising most rapidly at t = 15 s

The maximum rate of Jamie's rise is 2/3π m/s or 2.094 m/s.

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3 years ago