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2 years ago

10^(-10log3) How to solve this???

1 answer:

4 0

$a^{log_ab}=b\\and\\log_ab^c=c\cdot log_ab\\\\\\10^{-10log3}=10^{log3^{-10}}=3^{-10}\\\\\\3^{-10}=\frac{1}{3^{10}}=\frac{1}{59\ 049}$

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Svetradugi [14.3K]

Answer:

Step-by-step explanation:

(1, 5.5) ; (2, 9)

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$= \frac{9-5.5}{2-1}\\\\=\frac{3.5}{1}\\\\= 3.5$

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m = 7/2 ; (2 , 9)

y - y1 =m(x -x1)

$y - 9 = \frac{7}{2}(x- 2)\\\\y - 9 = \frac{7}{2}x - \frac{7}{2}*2\\\\y -9 = \frac{7}{2}x - 7\\\\y = \frac{7}{2}x - 7 + 9\\\\y = \frac{7}{2}x + 2$

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2 years ago

Actual lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy leng

HACTEHA [7]

Solution :

The data is normally distributed.

The standard deviation is 18 days

Here the data is normally distributed and 54 days is 3 days of standard deviation.

Therefore, the percentage of the births that would be $\text{expected to occur}$ within the 54 days of the mean $\text{pregnancy}$ length is given by :

= P( -3 < Z < 3)

= 0.9544

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Therefore, about 95% of the births would be $\text{expected to occur}$ within 54 days of the men $\text{pregnancy}$ length.

8 0

2 years ago

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butalik [34]

The answer would be 16 because 40 divided by 5 is 8 and 8 times 2 is sixteen so it would be 16

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3 years ago

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Korvikt [17]

-4 is greater than - 10 because -4 is closer to 0 than -10 is

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3 years ago

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olchik [2.2K]

Answer:

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Step-by-step explanation:

greatest common factor is 15cd

$15cd*1 + 15cd*2cd\\= 15cd(1 + 2cd)$

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1 year ago