Dividing two exponents with the same base has the effect of <em>subtracting</em> the bottom exponent from the top one. Following that rule, we can simplify our fraction like this:

$\dfrac{14^{-3}}{14^2}=14^{-3-2}=14^{-5}$

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3 years ago

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class size m average 0.82 n class n average 0.78. if the average for combined classes is 80%, what is relationship between m and

katen-ka-za [31]

Answer:

The relation between m and n is , m = n

Step-by-step explanation:

Given as :

The average for n class = 0.82 = 82%

The average for m class = 0.78 = 78%

The average for ( m + n ) class = 80%

So , $\frac{x_1 +x_2+x_3+x_4+ ....... + x_n}{n}$ = 82%

And $\frac{x_1 +x_2+x_3+x_4+ ....... + x_m}{m}$ = 78%

And , $\frac{(x_1+x_2+x_3+x_4+ ......+x_n ) + (x_1+x_2+x_+x_4+.....+x_m)}{m+n}$ = 80%

So , 0.82×n + 0.78×m = 0.80× (m+n)

Or, 0.82×n - 0.80×n = 0.80×m - 0.78×m

Or, 0.02×n = 0.02×m

Or, $\frac{m}{n}$ = $\frac{0.02}{0.02}$

∴ $\frac{m}{n}$ = 1

I.e m = n

Hence The relation between m and n is , m = n Answer

6 0

3 years ago