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Lesechka [4]
3 years ago
13

Suppose that you find the volume of all the oceans to be 1.4×109km3 in a reference book. To find the mass, you can use the densi

ty of water, also found in this reference book, but first you must convert the volume to cubic meters. What is this volume in cubic meters?
Physics
2 answers:
Licemer1 [7]3 years ago
8 0

Answer:

Explanation:

Volume of ocean = 1.4×10^9 km3

To m^3,

1.4 × 10^9 km^3 × (1000 m)^3/(1 km)^3

= 1.4 × 10^18 m^3.

Density of water = 1000 kg/m3

Mass = density × volume

= 1000 × 1.4 × 10^18

= 1.4 × 10^21 kg.

PSYCHO15rus [73]3 years ago
4 0

Answer:

V = 1.4\times 10^{18}\,m^{3},m = 1.435\times 10^{21}\,kg

Explanation:

The volume of all oceans is:

V = (1.4\times 10^{9}\,km^{3})\cdot (\frac{1\times 10^{9}\,m^{3}}{1\,km^{3}} )

V = 1.4\times 10^{18}\,m^{3}

Ocean water has a density of 1025\,\frac{kg}{m^{3}}, the mass of all oceans is:

m = (1025\,\frac{kg}{m^{3}} )\cdot (1.4\times 10^{18}\,m^{3})

m = 1.435\times 10^{21}\,kg

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Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

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The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

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a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

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b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

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= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

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Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

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