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Masja [62]
3 years ago
8

find the mean, median, and mode of he data set. round to the nearest tenth. 7, 5, 4, 2, 1, 1, 2, 3, 5, 4, 5

Mathematics
2 answers:
Ilya [14]3 years ago
5 0
Hello!

Mean: 7 + 5 + 4 + 2 + 1 + 1 + 2 + 3 + 5 + 4 + 5 = 39

 39 ÷ 11 = 3.5

The mean is 3.5

~~~~~~~~~~~~

Median: 1, 1, 2, 2, 3, 4, 4, 5, 5, 5, 7 - Which number is in the center?

The median is 4

~~~~~~~~~~~~~~~~~

Mode: Which number appears most often?

The mode is 5. 

Hope I helped! :3
SashulF [63]3 years ago
4 0
Hello!

First you need to list the numbers from least to greatest

1, 1, 2, 2, 3, 4, 4, 5, 5, 5, 7

To find the mean you add all the numbers up and divide the sum by the amount of numbers added

1 + 1 + 2 + 2 + 3 + 4 + 4 + 5 + 5 + 5 + 7 = 39

39 / 10 = 3.9

The mean is 3.9
------------------------------------------------------------------------------------------------------
To find the median you find the number in the middle

The median is 4
------------------------------------------------------------------------------------------------------
The mode is the number that appears the most

The mode is 5 

Hope this helps!
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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

6 0
3 years ago
Please help. I can not figure it out. I will mark brainliest.
zhuklara [117]
(0,30), (1,20)
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We know y intercept is 30
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6 0
2 years ago
HELP ASAP!!!!! HELP FAST!!!!!
mr_godi [17]

Answer:

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Step-by-step explanation:

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