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77julia77 [94]
2 years ago
6

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

rence of the Earth at the equator is 40,076 km.
Mathematics
1 answer:
Natalka [10]2 years ago
8 0

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}

Circumference in meters:

40,076km\times \frac{1000m}{1km}=40,076,000 m

Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

Hence, they encircle the planet 3.76\times 10^{11} times.

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Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

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(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2} * Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

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[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06

[(475.12-321.34) ± 2.485 *(33.06*\sqrt{\frac{1}{12} +\frac{1}{15} })]

[121.96; 185.60]

I hope this helps!

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2 years ago
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