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Mariulka [41]
3 years ago
5

The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{

kg})(kg)left parenthesis, start text, k, g, end text, right parenthesis. The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10 9 1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km 3 )left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10 12 km 3 kg ​ 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction . To the nearest hundredth of a percent, what percent of the earth's mass is from the oceans? (Ignore the \%%percent symbol when entering your answer. For example, if the answer is 0.11\%0.11%0, point, 11, percent, enter 0.110.110, point, 11.)
Mathematics
1 answer:
Alex_Xolod [135]3 years ago
6 0

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

 

km

3

kg

​

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

 

km

3

⋅1.03⋅10

12

 

km

3

kg

​

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

​

​

 

=

6⋅10

24

kg

1.3802⋅10

21

kg

​

=

6⋅10

3

1.3802

​

=

6000

1.3802

​

=0.0002300

3

​

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

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The Riemann sum with n = 6, taking the sample points to be midpoints is - 12.0625

<h3>What is Riemann sum?</h3>

Formula for midpoints is given as;

M = ∑0^n-1f((xk + xk + 1)/2) × Δx;

From the information given, we have the following parameters

  • x0 = 0
  • n = 6
  • xn = 3

Let' s find the parameters

Δx = (3 - 0)/6 = 0.5

xk = x0 + kΔx = 0.5k

xk+1 = x0 + (k +1)Δx

Substitute the values

= 0 + 0.5(k +1) = 0.5k - 0.5;(xk + xk+1)/2

We then have;

= (0.5k + 0.5k + 05.)/2

= 0.5k + 0.25.

Now f(x) = 2x^2 - 7

Let's find  f((xk + xk+1)/2)

Substitute the value of (xk + xk+1)/2)

= f(0.5k+ 0.25)

= 2(0.5k + 0.25)2 - 7

Put values into formula for midpoint

M = ∑05[(0.5k + 0.25)2 - 7] × 0.5.

To evaluate this sum, use command SUM(SEQ) from List menu.

M = - 12.0625

A Riemann sum represents an approximation of a region's area from addition of the areas of multiple simplified slices of the region.

Thus, the Riemann sum with n = 6, taking the sample points to be midpoints is - 12.0625

Learn more about Riemann sum here:

brainly.com/question/84388

#SPJ1

                                 

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