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2 years ago

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The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

has a density of 1.0 g/mL. How far has the boat sunk into the water?
Boat with cubes
A. 3.2 cm
B. 4.0 cm
C. 4.8 cm
D. 8.0 cm

1 answer:

zysi [14]2 years ago

7 0

Answer:

C

Step-by-step explanation:

:)

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A button machine produces one button in 0.15 seconds. How many buttons are produced in 48.6 seconds? 48.6 divided by 0.15 Multip

Kobotan [32]

Answer

Multiply the divisor and dividend by

✔ 100

The new division problem becomes

✔ 4,860 ÷ 15

The quotient of 48.6 ÷ 0.15 is

✔ 324

Estimating to check, you get

✔ 5,000 ÷ 15 = 333.33

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2 years ago

Read 2 more answers

Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway be

exis [7]

Answer:

a. k = 3

b. Cumulative distribution function X, $F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e. Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

$f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.$

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

$\int\limits^{\infty}_{-\infty} f(x)dx=1$

So, the equation becomes,

$\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3$

b. For this problem, the cumulative distribution function is defined as :

$F(x) = \int\limits^1_{\infty} f(x)dx + \int\limits^x_1 f(x)dx$

Now,

$F(x) = 0 + \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 + \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}$

Therefore the cumulative distribution function X is,

$F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

$P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125$

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

$P(2$

≅ 0.088

d. Mean value of headway,

$E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5$

And,

$E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3$

The standard deviation of headway is,

$= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254$

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

$P(\alpha - \beta < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X$

From part b, F(x) = 0, if x ≤ 1

$=1-(2.366)^{-3}\\=0.9245$

8 0

2 years ago

If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?

lana66690 [7]

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

$C = 9 * 10^2kg$

$C = 9 *100kg$

$C = 900kg$

$B = 1.8 * 10^5kg$

$B = 1.8 * 100000kg$

$B = 180000kg$

Divide the bigger weight by the smaller weight

$\frac{B}{C} = \frac{180000kg}{900kg}$

$\frac{B}{C} = \frac{180000}{900}$

$\frac{B}{C} = {200}{}$

Multiply both sides by C

$C * \frac{B}{C} = {200}{} * C$

$B = {200}{} * C$

$B = 200}C$

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

7 0

2 years ago

Kamila has a rectangular lawn that measures 80 by 150 feet. She wants to lay gravel made from colored rock in a border, 6 feet w

alex41 [277]

Answer:

1)Diagram attached

2)8424

3)1404

4)1569

Step-by-step explanation:

1)

Diagram is attached

2)

How many square feet of the gravel will surround the lawn:

Converting yards into feet:

1 yard =3 feet

So 80 yard =80(3) feet

=240 feet

and 150 yard =150 (3) feet

=450 feet

Area of rectangle is given by :length X width

Area of the outer border = (6+ 240+ 6 ) (6+ 450+6)

= 252 X 462

= 116424

Area of the lawn = 240 X 450

= 108000

Area of gravel border = 116424 - 108000

= 8424

Area of gravel border is 8424 ft^2

3)

Total volume when depth is 2 inches:

converting inches to feet:

1 inch= 0.083 feet

2 inches =0.1667 feet

volume = base area X depth

=8424 X 0.1667

= 1404

total volume of gravel needed for the border is 1404 ft^3

4)

Given:

A large bag of colored rock contains 5/8 of a cubic yard and costs $30

total cost of the rocks needed for the border=?

1 cubic yard = $30

converting cubic yard to cubic feet

1 cubic yard=27 cubic feet

cost of 1 cubic yard= cost of 27 cubic feet = $ 30

cost of 1 cubic feet=30/27

cost of 1404 cubic feet= 1404 X 30 / 27

= 1560

the total cost of the rocks needed for the border is $1560!

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kondaur [170]

75?i think not sureee I

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