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skad [1K]
3 years ago
13

Heather measures the temperature of her coffee to be 133.4 degrees fahrenheit. It is actually 145 degrees fahrenheit. What is th

e percent error of Heathers calculation?
Mathematics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

The percent error of Heather's calculation is <u>8%</u>.

Step-by-step explanation:

Given:

Heather measures the temperature of her coffee to be 133.4 degrees fahrenheit. It is actually 145 degrees fahrenheit.

Now, to find the percent error of Heather's calculation.

The temperature of coffee Heather measures = 133.4° F.

Coffee's actual temperature = 145° F.

So, to get the measurement in error we subtract the temperature of coffee Heather measures from coffee's actual temperature:

145-133.4=11.6\°.

Now, to get the percent error:

\frac{11.6}{145}\times 100

=\frac{1160}{145}

=8\%.

Therefore, the percent error of Heather's calculation is 8%.

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<em>x                    1/8            1/4             1/2              1             2</em>

<em>y                    -3                 -2            -1               0               1</em>

<em></em>

Step-by-step explanation:

<u><em>Explanation</em></u> :-

Given logarithmic function y = log_{b} (x)   if b >1

Given first table

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put x = \frac{1}{8}     given b > 1 so we can choose b = 2

y = log_{2} (\frac{1}{8} )

y = log_{2} (2^{-3}  )

we will apply logarithmic formula

log x ⁿ = n log (x)

y = log_{2} (2^{-3}  ) = -3 log_{2} (2) = -3 (1) = -3

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<em>ii)</em>

<em>put x = </em>\frac{1}{4}<em>     given b > 1 so we can choose b = 2</em>

<em></em>y = log_{2} (\frac{1}{4} )<em></em>

<em></em>y = log_{2} (2^{-2}  )<em></em>

we will apply logarithmic formula

log x ⁿ = n log (x)

y = log_{2} (2^{-2}  ) = -2 log_{2} (2) = -2 (1) = -2

<em>y = -2</em>

<em>iii) </em>

<em>put x = </em>\frac{1}{2}<em>     given b > 1 so we can choose b = 2</em>

<em></em>y = log_{2} (\frac{1}{2} )<em></em>

y = log_{2} (2^{-1}  )

<em>we will apply logarithmic formula </em>

<em>log x ⁿ = n log (x)</em>

y = log_{2} (2^{-1}  ) = -1 log_{2} (2) = - (1) = -1

<em>y = -1</em>

<em>iv) </em>

<em>put x = 1     given b > 1 so we can choose b = 2</em>

<em></em>y = log_{2} (1 )<em> = 0</em>

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<em>v) </em>

<em>put x = </em>2<em>     given b > 1 so we can choose b = 2</em>

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