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4 years ago

What is the solution to the equation below?3log_4x=log_43 2+log_4 2

1 answer:

6 0

$\frac{3\log _4\left(x\right)}{3}=\frac{\log _4\left(32\right)}{3}+\frac{\log _4\left(2\right)}{3}$
Simplify
$\log _4\left(x\right)=\frac{\log _4\left(32\right)+\log _4\left(2\right)}{3}$
$\log _4\left(x\right)=1$
Apply log rule
$\log _4\left(x\right)=\log _4\left(4\right)$
Answer:
$x=4$

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Answer:

-4x^4-6x^3-8x^2-24x

Step-by-step explanation:

$\left(2x^2+4x\right)\left(-2x^2+x-6\right)\\$

distribuir parênteses

$=2x^2\left(-2x^2\right)+2x^2x+2x^2\left(-6\right)+4x\left(-2x^2\right)+4xx+4x\left(-6\right)$

Aplicar regras de menos mais ; $+\left(-a\right)=-a$

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simplificar

$-2\times\:2x^2x^2+2x^2x-2\times\:6x^2-4\times\:2x^2x+4xx-4\times\:6x:\\\quad -4x^4-6x^3-8x^2-24x$

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4 years ago

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The 95% confidence interval for these parts is 56.98 to 57.05 under normal operations. A systematic sample is taken from the man

Nonamiya [84]

Answer:

Stop the line of production since the sample mean (56.96) is outside the confidence interval [ $CI_{0.95}$ = ( 56.98, 57.05 )]

Step-by-step explanation:

Given that;

95% confidence interval for these parts is 56.98 to 57.05

$CI_{0.95}$ = ( 56.98, 57.05 )

Sample mean = 56.96

We know that;

If the sample mean is within the confidence interval then, our decision is to keep the line operating as it is inside the confidence interval.

But if the sample mean is not within the confidence interval them, our decision is to stop the line of production as it is outside the confidence interval.

Now since our mean sample (56.96) does not lie between with the 95% confidence interval $CI_{0.95}$ = ( 56.98, 57.05 ).

Therefore, Stop the line of production since the sample mean (56.96) is outside the confidence interval [ $CI_{0.95}$ = ( 56.98, 57.05 )]

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3 years ago