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Makovka662 [10]
3 years ago
7

Here are the scores of students on a history test. 55, 56, 57, 58, 58, 64, 65, 66, 69, 75, 77, 78, 81, 81, 85, 86, 91. Create a

box and whisker plot

Mathematics
1 answer:
Anarel [89]3 years ago
6 0
The first thing you need to do is get the median of the whole set.

The median is the center so since there are 17 numbers the center is 69 and 69 is Q2

Next we need to find Q1 Q3 and Q4 Also, are whiskers are below.
Whisker minimum = 55
Whisker maximum = 91

Now to find Q1 we have to get the median of the first half of the data.
We need to get the median of the following data set.
55, 56, 57, 58, 58, 64, 65, 66

(58+58) / 2 = 58
Q1 = 58

Now we need to find Q3. To do that we get the median of the second half of the data.
75, 77, 78, 81, 81, 85, 86, 91. 
(81 + 81) / 2 = 81
Q3 = 81

Now we have all our quartiles and whiskers, which are below, it is time to create our box and whisker plot. Note, create a horizontal or vertical number line with appropriate data. Also the box part is from Q1 to Q3 and your whiskers goes from your minimum to your box and from your maximum to your box.

Whisker Minimum = 55
Q1 = 58
Q2 = 69
Q3 = 81
Whisker Maximum = 91

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Pamela Mello is paid on an incremental commission schedule. She is paid 2.6% on the first $70,000 and 3.4% on any sales over $70
Anarel [89]

Step-by-step explanation:

2.6% at $70,000

3.4% at $83,600 - $70,000

3.4% at $13,600

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5 0
3 years ago
At what point does she lose contact with the snowball and fly off at a tangent? That is
postnew [5]

Answer:

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight  down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the  instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to  the skier make with the vertical?"

- The figure is also attached.

Solution:

- The skier has a mass (m) and the snowball’s radius (r).

- Choose the center of the snowball to be the zero of gravitational  potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and  departs from the snowball.

- If we ignore snow-­ski friction along with air resistance, then the one work producing force in this problem, gravity,  is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical  energy at the top of the snowball.

- Hence, From conservation of energy we have:

                       KE (α) + PE(α) = KE(α = 0) + PE(α = 0)

                       0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r

                       0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r

                        m*v(α)^2 / r = 2*m*g( 1 - cos(α) )

- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity  can hold her to that path, i.e. as long as:

                         m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal  force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

                            3*cos(α) ≥ 2

                            α ≥ arc cos ( 2/3) ≥ 48.2°

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3 years ago
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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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