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3 years ago

Solve to indicated variable 16ak=-5 for a

1 answer:

3 0

The answer is
$\frac{ - 5}{16k} = a$
steps
#1) Try to leave a by itself by dividing any variable next to it
$\frac{16 ak = - 5 }{ 16k}$
you can only divided these variables as all these variables in the original equation. are being multiplied. The opposite if multiplication is division.

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2 years ago

There is 3/4 of a box of cereal remaining. You eat 2/5 of the remaining cereal. What fraction of the box do you eat? Hint: What

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2 years ago

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3 years ago

Solve 5(x-1)^1/3 = 20

dangina [55]

Answer:

x = 65

Step-by-step explanation:

To solve the equation, we write it in radical form:

$5\, \sqrt[3]{x-1} = 20$

Then, we divide both sides by 5 to isolate the root:

$\sqrt[3]{x-1} = \frac{20}{5} \\\sqrt[3]{x-1} =4$

and next, we raise both sides of the equation to the power 3 so as to free the expression in x from the root:

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now we add 1 to both sides to isolate x:

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8 0

2 years ago

(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the

goblinko [34]

Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

$98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\ \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\ \\ =10.0023\pm2.33(0.000089) \\ \\ =10.0023\pm0.00021 \\ \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\ \\ =(10.0021, \ 10.0025)$

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.

Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

$\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\ \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\ \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\ \\ \Rightarrow n=4.66^2=21.7156\approx22$

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>

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3 years ago