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ratelena [41]
3 years ago
9

Period 1 has 15 students in it and a test average of 86%. Period 2 has 21 students in it and an average of 88%. Period 3 has 12

students in it and an average of 95%. Use weighted means to find the overall average of the classes
Mathematics
2 answers:
Tanzania [10]3 years ago
8 0

Answer:

The overall average of the classes is 89.125%

Step-by-step explanation:

For calculate the average of the class using weighted means, it is necessary to take into account the number of students of every period. So, the average can be calculated as:

Average =\frac{(x_1*a_1) +( x_2*a_2) + (x_3*a_3)}{x_1 + x_2 + x_3}

Where isx_1, x_2 and x_3 are the number of students for period 1 , 2 and 3 respectively and a_1, a_2 and a_3 are the average of period 1, 2 and 3 respectively.

Replacing the values of x1 by 15 students, x2 by 21 students, x3 by 12 students, a1 by 86%, a2 by 88% and a3 by 95%, we get:

Average = \frac{(15*86)+(21*88)+(12*95)}{15+21+12}

Average = 89.125%

So, using weighted means, the overall average of the classes is 89.125%

STatiana [176]3 years ago
7 0
Add all of the percents up and then divide by 3. You get 89.6% 
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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?
astraxan [27]
ANSWER


The value of the expression is
- 1


EXPLANATION

Method 1: Rewrite as product of
{i}^{2}


The expression given to us is,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}


We use the fact that
{i}^{2}  =  - 1
to simplify the above expression.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


We substitute to obtain,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  1 \times   1  \times  - 1 =  - 1


Method 2: Use indices to solve.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{0 + 1 + 2 + 3 + 4}



This implies that,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{10}




{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (  {{i}^{2}} )^{5}


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


8 0
4 years ago
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