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2 years ago

Solve for b b^-4=1/81^=exponent -4

Mathematics

2 answers:

AlekseyPX2 years ago

3 0

$b^{-4}= \frac{1}{81} \\ \\ \frac{1}{b^4} = \frac{1}{81} => \\ \\ b^4=81 \\ b^4=3^4=>b=3
$

g100num [7]2 years ago

3 0

1 / ( b^ 4 ) = 1 / 81 ;
b^4 = 81 ;
b^4 = 3^4 ;
b = + 3 or b = - 3 ;

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LuckyWell [14K]

Answer:

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Step-by-step explanation:

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$a_n=a_1x^{n-1}$

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The equations are very similar and you only really need to rearrange it. The factor (2/3) and the first term (9) are given, so you can write the iterative equation:

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1 year ago

The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.

nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

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Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

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With reference to table we get

$\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418$

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85$

and

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97$

∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31

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3 years ago

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olga nikolaevna [1]

Step-by-step explanation:

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Likurg_2 [28]

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2 years ago