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alina1380 [7]
2 years ago
9

In the first event, the eighth graders are running a baton relay race with three other classmates. The team’s top speed for each

leg is 56.81 seconds, 59.22 seconds, 57.39 seconds, and 60.11 seconds. Use the information to predict the team’s best time for the race.
Mathematics
2 answers:
anzhelika [568]2 years ago
4 0

Answer:

56.81 seconds is the team’s best time for the race

Step-by-step explanation:

Here, we want to give the team’s best time for the race

from the question, we should understand that the best times are the smallest in values since it is a race

The smallest time here is 56.81 seconds and thus that is the team’s best time for the race

Simora [160]2 years ago
4 0

Answer:

He’s right I checked

Step-by-step explanation:

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He has a total of 40

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he has a total of $20 left

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
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Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
Ella and her children went into a bakery and where they sell cookies for $0.50 each and brownies for $2.25 each. Ella has $15 to
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$15 - $2.5 = $12.5

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