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andreyandreev [35.5K]
2 years ago
8

Which graph represents the function f (x) = startfraction 4 over x endfraction?.

SAT
1 answer:
Fynjy0 [20]2 years ago
3 0

The graph of the function is plotted below

<h3 /><h3>Graphs and functions</h3>

Graph of functions is plotted on the xy-plane. The graph can be a curve or a line depending on the nature of the function.

<h3>Linear and polynomial graphs</h3>

Most linear equations are represented by a straight line while curves can be an exponential or polynomial function.

According to the question, we are to plot the function f(x) = \dfrac{4}{x}. The equivalent graph of the function is as shown in the attachment.

Learn more on graphs here: brainly.com/question/4291574

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Answer:

0.2776

Explanation:

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We need to find Cumulative probability: P(X > 15884)

First we need to convert it into normal distribution.

From the attached file, we can see the shaded area we are looking for.

For conversion as follow; P (X>15884) =P ( X - mean > 15884-15000 ) =P (\frac{X- mean}{SD} > \frac{15884-15000}{1500} )

Next to find Z = \frac{X- mean}{SD} = 0.59

now we have converted to normal distribution and found the value of Z, we need to find the probability P(X > 15884).

P(X > 15884) = P(Z > 0.59).

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When the chemical reaction 2no(g)+o2(g)→2no2(g) is carried out under certain conditions, the rate of disappearance of no(g) is 5
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The rate of disappearance of O2(g) under the same conditions is 2.5 × 10⁻⁵ m s⁻¹.

<h3>What is the rate law of a chemical equation? </h3>

The rate law of a chemical reaction equation is usually dependent on the concentration of the reactant species in the equation.

The chemical reaction given is;

\mathbf{2 NO_{(g)} + O_{2(g)} \to 2 NO_{2(g)} }

The rate law for this reaction can be expressed as:

\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}= +\dfrac{1}{2}\dfrac{d[NO_2]}{dt}}

Recall that:

  • The rate of disappearance of NO(g) = 5.0× 10⁻⁵ m s⁻¹.

  • Since both NO and O2 are the reacting species;

Then:

  • The rate of  disappearance of NO(g) is equal to the rate of  disappearance of O2(g)

\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}}

\mathbf{= -\dfrac{1}{2} \times 5.0 \times 10^{-5}  = rate \  of  \ disappearance \ of  \ O_2}

Thus;

The rate of disappearance of O2 = 2.5 × 10⁻⁵ m s⁻¹.

Therefore, we can conclude that two molecules of NO are consumed per one molecule of O2.

Learn more about the rate law here:

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