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Igoryamba
3 years ago
9

Derivative of sin (x^(1/3)) + (sinx)^(1/3)

Mathematics
2 answers:
Strike441 [17]3 years ago
8 0
1/3 x^(-2/3) cos (x^1/3) + 1/3(sinx)^-2/3 .cos x



iris [78.8K]3 years ago
3 0
\bf sin\left( x^{\frac{1}{3}} \right)+[sin(x)]^{\frac{1}{3}}\\\\
-------------------------------\\\\
cos\left( x^{\frac{1}{3}} \right)\cdot \cfrac{1}{3}x^{-\frac{2}{3}}+\cfrac{1}{3}[sin(x)]^{-\frac{2}{3}}\cdot cos(x)\impliedby \textit{chain-ruling both terms}
\\\\\\
\cfrac{cos(\sqrt[3]{x})}{3\sqrt[3]{x^2}}+\cfrac{cos(x)}{3\sqrt[3]{sin^2(x)}}
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Step-by-step explanation:

Move the decimal place from 2.375 to the back of the number so it should look like this, 2375.

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Anuta_ua [19.1K]

Answer:

The confidence Interval = (0.329, 0.391)

Step-by-step explanation:

Formula for the Confidence Interval for proportion =

p ± z × √p(1 - p)/n

where

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From the question

n = 900 people

x = 36% of 900 people

= 36/100 × 900

= 324

z = z score of 95% confidence Interval

z = 1.98

p = x/n

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= 0.36 ± 1.96 × √0.36 (1 - 0.36)/900

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