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koban [17]
3 years ago
14

Earth's oceans have an average depth of 3800 m, a total area of 3.63 × 108 km2, and an average concentration of dissolved gold o

f 5.8 × 10−9 g/L. If a recent price of gold was $129.00/troy oz, what is the value of gold in the oceans?
Mathematics
1 answer:
Phoenix [80]3 years ago
3 0

Answer:

value of gold is $33.17

Step-by-step explanation:

given data

depth d = 3800 m = 3.8 km

area A  = 3.63 × 10^8 km²

average concentration = 5.8 × 10^−9 g/L

price of gold = $129.00 / troy oz

to find out

value of gold

solution

first we find volume that is

volume v = A× d

put here value

volume = 3.8 ×  3.63 × 10^8

volume =  13.794 × 10^{8} km³

volume = 137.94 × 10^{19} L

so mass =  5.8 × 10^{9} × 137.94 × 10^{19}

mass = 8 × 10^{12} g

mass = 0.2572 troy oz

here we know 1 troy oz = 31.1 g

and we know 1 troy oz = $129

so 0.2572 troy oz = 129 × 0.2572

0.2572 troy oz value is $33.17

value of gold is $33.17

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