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blagie [28]
3 years ago
13

I need these THREE questions ANSWERED please!!!! please i really need these done!!!

Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0

Step-by-step explanation:

8. As the name implies, a box-and-whisker plot looks like a box with two whiskers.  The box is the middle 50%.  The whiskers are the bottom 25% and the top 25%.

First, we need to sort the numbers from smallest to largest.  Starting with Set A:

Set A = {56, 57, 62, 68, 71, 82, 84, 92, 97, 101, 103, 106}

Now we find the median of the set, or the middle number.  Set A has 12 data points.  Since 12 is an even number, the median will be the average of the 6th and 7th numbers.

M = (82 + 84) / 2 = 83

Next, we find the median of the lower half.  There are 6 numbers in the lower half, so the median is the average of the 3rd and 4th numbers.

Ml = (62 + 68) / 2 = 65

Now we find the median of the upper half.  Again, there are 6 numbers in the upper half, so the median will be the average of the 3rd and 4th numbers.

Mu = (97 + 101) / 2 = 99

So the left whisker is from 56 to 65.

The box is from 65 to 99.

The right whisker is from 99 to 106.

Don't forget to mark the median, 83.

Repeat for Set B.

Set B = {36, 37, 42, 46, 48, 56, 58, 63, 69, 72, 75, 78}

M = (56 + 58) / 2 = 57

Ml = (42 + 46) / 2 = 44

Mu = (69 + 72) / 2 = 70.5

So the left whisker is from 36 to 44.

The box is from 44 to 70.5.

The right whisker is from 70.5 to 78.

The median is 57.

By graphing both on the same number line, we can easily compare them.

9. To make a dot plot, draw a number line starting from the smallest number and ending at the largest number.  For each number, draw a dot every time the number appears in the set.  For example, the number 3 appears twice in Set A, so draw two dots on 3.  You may find it helpful to sort the data first.

10. Draw a dot plot of the set to determine the shape.

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Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​
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<u>Prove that:</u>

\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0

<u>Proof: </u>

We know that, by Law of Cosines,

  • \sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}
  • \sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}
  • \sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}

<u>Taking</u><u> </u><u>LHS</u>

\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>On combining the fractions,</em>

\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>Regrouping the terms,</em>

\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}

\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}

\longmapsto\dfrac{0}{2abc}

\longmapsto\bf 0=RHS

LHS = RHS proved.

7 0
3 years ago
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