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ioda
3 years ago
6

The owner of a fish market has an assistant who has determined that the weights of catfish are normally​ distributed, with mean

of 3.2 pounds and standard deviation of 0.8 pound. what percentage of samples of 4 fish will have sample means between 3.0 and 4.0​ pounds
Mathematics
1 answer:
wel3 years ago
8 0
<span>21.72%
The 68-95-99 rule states that 1 standard deviation from the mean will contain 68% (68.27 more specifically) of all data points. For each sample, we multiply the success rate for </span>.6827*.6827*.6827*.6827 =.2172 or 21.72%
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Answer:

\sf \log_{10}6=0.7781

Step-by-step explanation:

<u>Given</u>:

\sf \log_{10} 2 = 0.3010

\sf \log_{10} 3 = 0.4771

To find log₁₀ 6, first rewrite 6 as 3 · 2:

\sf \implies \log_{10}6=\log_{10}(3 \cdot 2)

\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay

\implies \sf \log_{10}(3 \cdot 2)=\log_{10}3+\log_{10}2

Substituting the given values for log₁₀ 3 and log₁₀ 2:

\begin{aligned} \sf \implies \log_{10}3+\log_{10}2 & = \sf 0.4771+0.3010\\ & = \sf 0.7781 \end{aligned}

Therefore:

\sf \log_{10}6=0.7781

Learn more about logarithm laws here:

brainly.com/question/27953288

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