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lukranit [14]
2 years ago
14

Class a had a mean score of 7.8 and a standard deviation of 1 class b had a mean score of 8.1 and a standard deviation of 0.2 wh

ich class scored better on average?
Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Class 2 scored better on average

<h3>How to determine which class scored better on average?</h3>

The averages and the standard deviations are given as:

Class 1

Mean score = 7.8

Standard deviation = 1

Class 2

Mean score = 8.1

Standard deviation = 0.2

As a general rule, the smaller the standard deviation; the better the average

Class 2 has a smaller deviation

Hence, class 2 scored better on average

Read more about standard deviation at:

brainly.com/question/28061243

#SPJ1

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Anybody help me to solve this question. ​
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Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

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