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7 days ago

4. Odette Alverez works at the Digitech Warehouse eight hours a day, earning $9.48 an hour. She is asked to work two additional

hours at time and a half to help prepare for an incoming shipment of merchandise. How much will she earn for these two hours of overtime?

Mathematics

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A bag holds 12 red marbles, 11 green marbles, 17 blue marbles, and 5 yellow marbles. What is the probability that you will not c

Zina [12379]

¡Hola! Bienvenido a!

Vamos a sumar cuántas canicas tenemos en total.

12+11+17+5=45

Queremos hallar la probabilidad de elegir una canica que no sea azul. Observemos cuántas canicas no son azules.

12+11+5=28

Tendremos esta probabilidad sobre 48.

28/48

Al simplificar, obtenemos 7/12 o alrededor de 58.33%.

¡Espero que esto ayude!

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3 months ago

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The fish population of Lake Collins is decreasing at a rate of 3% per year. In 2004 there were about 1,300 fish. Write an expone

Inessa [12570]

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2 months ago

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The diagonals of the figure below represent the support beams for a patio covering.

Svet_ta [12734]

The response is B: 5 and 5 square root 3.

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2 months ago

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

2 months ago