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14 days ago

Paul is going to have the exterior of his home painted. Painting USA charges $75.00 to come out and evaluate the house plus $25.

00 per hour of labor. Upscale Home Painting does not charge to evaluate the house. However, they charge $40.00 per hour. If it takes 25 hours to paint Paul’s house, the most cost effective company will be?The cost of painting the house will be the same for both companies if the number of hours required to paint Paul’s house is?

Mathematics

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the diameter of a cylindrical box of CDs is 5 inches. The height of the box is 4 inches. Whuch measurement is closest to the tot

Svet_ta [12734]

To calculate the area, simply multiply 5 by 4, resulting in 20.

5 0

2 months ago

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The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

AnnZ [12381]

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

La solución de la integral de la ecuación (1) es:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

Espero que te ayude.

3 0

3 months ago

There are seven boys and five girls in a class. The teacher randomly selects three different students to answer questions. The f

Svet_ta [12734]

The likelihood of selecting one girl is calculated as $\frac{5}{12}$ . This is based on having 5 girls within a total of 12 students, and the probability of an event can be expressed as: $\frac{\text{# of things you want}}{\text{# of things are possible}}$ .

Using the same reasoning, for the next student, we have reduced the number of students by 1, leading to 11 possible outcomes instead of 12, giving us: $\frac{7}{11}$ , which represents the probability of selecting a boy as the second choice.

Lastly, the probability of choosing a girl for the third selection follows the same logic and is given as: $\frac{4}{10}$ .

However, we must combine these individual probabilities to determine the likelihood of this specific sequence of selections occurring:

$\frac{5}{12}*\frac{7}{11}*\frac{4}{10}=\frac{140}{1320}$

This simplifies to:

$\frac{7}{66}$

4 0

4 months ago

Read 2 more answers