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Georgia [21]
3 years ago
10

Ratio, Rate, and Speed

Mathematics
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

10:12:15

Step-by-step explanation:

a/b=5/6

b/c=4/5

thus

c/b=5/4

LCM of 4 and 6=12

5/6=10/12

5/4=15/12

b is the 12 in both

thus ratio between a:b:c is 10:12:15

i can reply in comments if more explanation is needed

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The manager of a grocery store selected 100 customers at random to ask what the maximum price is that he or she would pay for a
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Maybe C, but I'm not sure, so you should check :)
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3 years ago
The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{
Alex_Xolod [135]

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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To the right: 0.51
to the left: -0.51
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Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En
ArbitrLikvidat [17]

<em>Y</em>₁ and <em>Y</em>₂ are independent, so their joint density is

f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}

By definition of conditional probability,

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = P((<em>Y</em>₁ > <em>Y</em>₂) and (<em>Y</em>₁ < 2 <em>Y</em>₂)) / P(<em>Y</em>₁ < 2 <em>Y</em>₂)

Use the joint density to compute the component probabilities:

• numerator:

P((Y_1>Y_2)\text{ and }(Y_1

=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1

=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1

=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1

=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty

=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16

• denominator:

P(Y_1

(I leave the details of the second integral to you)

Then you should end up with

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = (1/6) / (2/3) = 1/4

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Answer:

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Step-by-step explanation:

look at the picture

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3 years ago
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