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zhenek [66]
3 years ago
14

A 17-inch candle is lit and burns at a constant rate of 1.3 inches per hour. Let t represent the number of hours since the candl

e was lit, and suppose f is a function such that f ( t ) represents the remaining length of the candle (in inches) t hours after it was lit. Write a function formula for f . f ( t )
Mathematics
1 answer:
pshichka [43]3 years ago
6 0

Answer

Since the lyla deleted it actually for a good reason let me explain. But you dont just delete random answers, you can give a comment. I can report you for doing that.

So since it burns at 1.3 per hour. Notice that since it gets removed per hour. so it would be 1.3t so it multiplies per hour. Then we need to subtract. Thats because it lowers the length of the candle. So we get the original length - 17 and get the function

f(t) = 17-1.3t

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Your car invoice reads: “Price $1,492.50, dealer preparation $45.00, transportation $54.50, undercoat $148.80, 60-day guarantee
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3 years ago
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Answer:

a) <em> standard error of the mean =10.06</em>

<em>b)  The margin of error  = 17.3982</em>

<em>c) 95% of confidence intervals are </em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d) Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

<u>Step-by-step explanation:</u>

<u><em>Step(i)</em></u><u>:-</u>

Given sample size 'n' = 20

Given sample mean was found to be 107 bpm with a standard deviation of 45 bpm.

<em>Sample mean             </em>x^{-} = 107 bpm<em></em>

<em>Sample standard deviation (S) = 45 bpm</em>

<em>a) standard error of the mean is determined by</em>

<em>     </em>S.E = \frac{S}{\sqrt{n} } = \frac{45}{\sqrt{20} }<em></em>

<em>     S.E = 10.06</em>

<em>b) The margin of error is determined by</em>

<em></em>M.E = \frac{t_{\alpha } X S }{\sqrt{n} }<em></em>

<em>The degrees of freedom  ν   </em>= n-1 =20-1=19<em></em>

<em>   </em>t_{\alpha } = 1.729  

<em></em>M.E = \frac{1.729X 45}{\sqrt{20} }<em></em>

<em></em>M.E = \frac{77.805 }{4.472 } = 17.3982<em></em>

<em>c) 95% of confidence intervals are determined by</em>

<em></em>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} + t_{\alpha } \frac{S}{\sqrt{n} } )<em></em>

<em></em>(107 -  1.729\frac{45}{\sqrt{20} } , 107 + 1.729\frac{45}{\sqrt{20} } )<em></em>

<em></em>(107 -  17.3982 } , 107 +17.3982 )<em></em>

<em></em>(89.6018 ,124.3982)<em></em>

<em>d)  </em>

<em>Lower limit of 95% of confidence interval  = 89.6018</em>

<em>upper limit of 95% of confidence interval  = 124.3982</em>

<em>The Population mean is lies between in these intervals</em>

<em></em>

<em></em>

4 0
3 years ago
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