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2 years ago

A group of 17 students participated in a quiz competition. Their scores are shown below:

Mathematics

1 answer:

DochEvi [55]2 years ago

3 0

Part A

<h3>Answer: Histogram</h3>

Explanation:

A histogram is best because it helps show how the scores breakdown for the various bins or intervals or classes. It's effectively a visual summary of a grouped frequency table.

A dot plot would not work because we don't know the specific scores, but rather just the general region of where they fall.

A box plot doesn't work well either because we lose information on how the distribution is broken down. Sure we get the five number summary out of it, but we don't see how the bins are set up.

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Part B

We have the categories or bins of 0-4, 5-9, 10-14, 15-19, 20-24. Each bin will correspond to a bar that tells us how frequently the values show up. It tells us how frequent or how many scores we got in that sub-interval.

The bin 0-4 has a bar that is 4 units high which means we got 4 scores that are between 0 and 4
The bin 5-9 has a bar that is 5 units high, so we have 5 scores between 5 and 9.
The bin 10-14 has a bar that is 2 units high since we have 2 scores in this region.

And so on. This all leads to the histogram shown below.

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Answer:

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Which is option C

Step-by-step explanation:

Given:

Slope = m = -2

Given points are ( -2,12)

Which is

x = -2 and y =12

TO find:

Equation of line passing through these points = ?

Solution:

The point slope form of a line is

y = mx + c

Here we don't know the value of c

To find it

Putting y = 12 , x= -2 and y = -2 in the given equation

y = mx + c

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12 = (-2)*(-2) + c

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Subtracting 4 from both sides

12-4 = 4 -4 + c

8 = c

Now we have

m = -2 and c= 8

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y = mx + c

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y = -2*x + 8

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3 years ago

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Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?

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Lets Find It Out..

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As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
→n=<1,2−2>

The equation of the plane parallel to the original one passing through P(x0,y0,z0)is:

→n⋅< x−x0,y−y0,z−z0>=0
<1,2,−2>⋅<x−x0,y−y0,z−z0>=0
x−x0+2y−2y0−2z+2z0=0
x+2y−2z−x0−2y0+2z0=0

x+2y−2z+d=0 [1]
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Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when x=0 and y=0:
x+2y−2z=1 => 0+2⋅0−2z=1 => z=−12
→P1(0,0,−12)

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping D=2, and d as d itself, we get:

D=|ax1+by1+cz1+d|√a2+b2+c2
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d+1=−6 => d=−7
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