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goblinko [34]
3 years ago
13

How do you solve st+3t=6

Mathematics
1 answer:
bonufazy [111]3 years ago
4 0
You will factor out the common factor of t if you are solving for t. you will then divide both sides by (s+3) to make t the subject of the formula. for this question you can only get t in terms of s.
so:
st+3t=6
t(s+3)/(s+3)=6/(s+3)
t=6/(s+3)

if you are required to solve for s then you are going to subtract 3t from both sides. you can factorise 6-3t .you will then divide both sides by t to isolate s.for this question you can only get s in terms of t.so:

st+3t-3t=6-3t
st/t=2(3-t)/t
s=2(3-t)/t
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Answer:

1715 hours

Step-by-step explanation:

7hours ×5 days= 35 horus a week

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5 0
4 years ago
Combine like terms:<br> 7x + 3x + 4y - X+.5y
Murrr4er [49]

Answer:

9x + 4.5y if .5y is a decimal point, 9x + 9y if the dot is a typo

Step-by-step explanation:

If the dot is real,

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If the dot is a typo,

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7 0
3 years ago
Read 2 more answers
(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>
8 0
4 years ago
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IgorLugansk [536]

Answer:

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Step-by-step explanation:

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Iteru [2.4K]

Answer:

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Step-by-step explanation:

Hope this helps!

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