Solve log 3x -log 13=2. Round to the nearest thousandth if necessary.

1 answer:

4 0

$\bf \textit{Logarithm of rationals}\\\\
log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y)
\\\\\\
\textit{Logarithm Cancellation Rules}\\\\
log_a a^x= x\qquad \qquad \boxed{a^{log_ax}=x}\\\\
-------------------------------\\\\
log(3x)-log(13)=2\implies log_{10}(3x)-log_{10}(13)=2
\\\\\\
log_{10}\left( \frac{3x}{13} \right)=2\implies 10^{log_{10}\left( \frac{3x}{13} \right)}=10^2\implies \cfrac{3x}{13}=2
\\\\\\
3x=26\implies x=\cfrac{26}{3}\implies x=8\frac{2}{3}$

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The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What ar

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Answer:

The minimum value of the bill that is greater than 95% of the bills is $37.87.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 28, \sigma = 6$