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3 years ago

5

a student designed an experiment to show how water is recycled through the atmosphere. which of these is least likely to be a pa

rt of the experiment? a burner to heat water. a beaker in which water is boiled. a cup of ice cubes to cool the water. a cold glass plate on which water vapor condenses.

2 answers:

exis [7]3 years ago

5 0

<span>a cup of ice cubes to cool the water.</span>

daser333 [38]3 years ago

4 0

Answer:

The answer is - a cup of ice cubes to cool the water.

Step-by-step explanation:

The thing that is least likely to be a part of the experiment is - a cup of ice cubes to cool the water.

A burner to heat water and a beaker in which water is boiled. These are essential as warming water will result in generating vapor. And these vapors can be condensed on a cold glass plate, as water vapor condenses easily on cold surfaces.

So, the unusable thing here is - a cup of ice cubes to cool the water.

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What is the term to this?

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3 years ago

X-2y=18 2x+y=6 parallel perpendicular or neither

sergey [27]

<h3>Therefore they are perpendicular.</h3>

Step-by-step explanation:

A equation of line is

y =mx +c

Here the slope of the line is m.

Given equations are

x - 2y = 18

⇔-2y = -x +18

$\Leftrightarrow y =\frac{1}{2} x -9$ ............(1)

and 2x + y = 6

⇔y = -2x +6 ............(2)

Therefore the slope of equation (1) is $(m_1)$ = $\frac{1}{2}$

Therefore the slope of equation (2) is $(m_2)$ = -2

If two lines are perpendicular, when we multiply their slope we get -1.

therefore,

$m_1. m_2$ = $\frac{1}{2}$ . (-2) = -1

Therefore they are perpendicular.

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3 years ago

What is standard form

nataly862011 [7]

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expanded form:
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standard form:
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3 years ago

Read 2 more answers

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

$f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.$

The value of constant C can be obtained as:

$\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1$

$\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1$

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$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1$

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$10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1$

$50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1$

$50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1$

$50C[0+\frac{1}{0.5} ] =1$

$100C = 1$ ⇒ $C = \frac{1}{100}$

C = 0.01

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