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3 years ago

Pls help me solve this problem!!

Mathematics

1 answer:

ddd [48]3 years ago

4 0

Answer:

49.85

Step-by-step explanation:

41 is the mean, so half of the daily requests are above that number, half below, the question is about numbers above the mean, so the lower numbers won't be included in the percentage.

The difference between 59 and 41 is 18. The standard deviation given is 6.

18 ÷ 6 = 3 So that is 3 standard deviations. The empirical rule states that 99.7 of all values are within 3 standard deviations from the mean. But we are looking at only the upper half of those values so 99.7 ÷ 2 = 49.85 %

If the answer asks for approximate, you could round to 49.9 or 50.

And tell the math people to learn the correct spelling of fluorescent!

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Answer:

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Step-by-step explanation:

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Which number produces a rational number when added to 0.25

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1 year ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Aleonysh [2.5K]

Answer:

t=5.18 minute

Step-by-step explanation:

Using Newton's law of cooling

$\frac{dT}{dt}=-k(T-T_{s})$

T is temperature of a cup of coffee at any time t $T_{s}$ , is the temperature of the surrounding and k is a constant of proportionality in this negative because temperature is decreasing.

$T(0)=190\\T(1)=170\\T_{s}=66$

$\frac{dT}{dt}=-k(T-T_{s})\\\frac{dT}{(T-T_{s})}=-k*dt\\\int\limits {\frac{1}{T-T_{s} } } \, dT=-\int\limits {k} \, dt\\ln(T-T_{s})=k*t+C\\e*ln(T-T_{s})=e^{k*t+C} \\T-T_{s}=e^{k*t}*e^{C}\\e^{C}=C\\T-T_{s}=e^{k*t}*C\\T=T_{s}+e^{k*t}*C$

To find constant knowing the time and the temperature the first step of the change of energy in cup of coffee

$T=T_{s} +e^{-k*t}*C\\ C=190-66=124\\170=66+124*e^{-k*2}\\ 170-66=124*e^{-k*2}\\ln(\frac{104}{124})=ln*e^{-k*2}\\-0.175=-k*2\\k=0.08794$

Now using the constant of decreasing can find the time to be a 145 temperature the cup of coffee

$145=66+124*e^{-k*t} \\145-66=124*^{-0.087*t}$

$ln(0.637)=ln*e(-0.087*t)\\-0.45=-0.087*t\\t=5.18minutes$

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3 years ago