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2 years ago

Please help me

Mathematics

2 answers:

timurjin [86]2 years ago

6 0

$3(4x-2)=2(6x-3) \\ \\ 12x - 6 = 12x - 6 \\ \\ Answer: \fbox {Infinite Solutions}$

dusya [7]2 years ago

4 0

3(4x - 2) = 2(6x - 3)

12x - 6 = 12x - 6

Add 6 on both sides

12x = 12x

Divide by 12 on both sides

x = x

That means this has infinite solutions. Your answer is B.

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2 years ago

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Help me pls I'm so confused and this is due in the morning pls help. look at pic

Sloan [31]

Answer:

the equation that has the same solution as the given expression is

m + 7.28 = 18.28

Step-by-step explanation:

Given expression;

0.2m = 2.20

To solve for m, divide both sides by 0.2

$\frac{0.2 m}{0.2} = \frac{2.2}{0.2} \\\\m = 11$

From the given options, the equation that has the same solution as the given expression is m + 7.28 = 18.28

Solve the equation to confirm as follows;

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collect like terms

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2 years ago

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photoshop1234 [79]

Answer:

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Step-by-step explanation:

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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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