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AveGali [126]
3 years ago
10

Please help! It’s Saturday and I only have this question left on my digital worksheet!

Mathematics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

<h2>The answer is option C</h2>

Step-by-step explanation:

<h3>\frac{ {6}^{ - 3} }{ {6}^{5} }</h3>

Using the rules of indices

Since the bases are the same and are dividing we subtract the exponents

That's

<h3>\frac{ {a}^{x} }{ {a}^{y} }  =  {a}^{x - y}</h3>

So we have

<h3>\frac{ {6}^{ - 3} }{ {6}^{5} }  =  {6}^{ -  3 - 5}  =  {6}^{ - 8}</h3>

Using the rules of indices

<h3>{x}^{ - y}  =  \frac{1}{ {x}^{y} }</h3>

So we have the final answer as

<h2>\frac{1}{ {6}^{8} }</h2>

Hope this helps you

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Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

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3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

D

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