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vesna_86 [32]
3 years ago
14

(I will mark brainliest if you can get it correct) Hot chocolate is made with milk warmed to 172∘F with an error tolerance of 6%

. What is the acceptable temperature range for milk when making hot chocolate?
A. 170.968∘F to 173.032∘F


B. 161.68∘F to 182.32∘F


C. 171.4∘F to 172.6∘F


D. 166∘F to 178∘F
Mathematics
2 answers:
laiz [17]3 years ago
7 0

Answer:

161.68°F to 182.32°F

Step-by-step explanation:

6%=6/100= 0.06

0.06×172=10.32°F

thus temperature range is

(172 + or - 10.32)°F

tangare [24]3 years ago
5 0
D cuz I feel like it is
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Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13

x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88

x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63

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x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13

So, the table becomes:

\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}

Next, calculate the mean

\bar x = \frac{\sum fx}{\sum f}

\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}

\bar x = \frac{1845.73}{146}

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Next, the sample variance is:

\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}

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\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}

\sigma^2 = \frac{2393.6875}{145}

\sigma^2 = 16.51

The sample standard deviation is:

\sigma = \sqrt{\sigma^2}

\sigma = \sqrt{16.51}

\sigma = 4.06

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Step-by-step explanation:


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