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1 year ago

I don’t have a question because I posted a photoz

Physics

1 answer:

Scorpion4ik [409]1 year ago

7 0

Part a:

$Q_{1}$ = 56

$Q_{2}$ = 60

$Q_{3}$ = 63

The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The $Q_{2}$ is the overall medium, $Q_{1}$ is the medium of the first half, and $Q_{3}$ is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between $Q_{1}$ and $Q_{3}$ . In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + $\frac{7}{2}$ ) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

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vekshin1

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$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

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Here, $T_m$ is the air temperture, C and k are constants.

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$70^\circ F$ in $t=0$

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$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

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