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19 days ago

If you know that a < b, and both a and b are positive numbers, then what must be true about the relationship between the oppo

sites of these numbers? Explain.

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Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a per

lawyer [12517]

Response: (0.8115, 0.8645)

Step-by-step outline:

Define p as the proportion of individuals who leave one space after a sentence.

Provided: Sample size: n= 525

Number of respondents indicating they leave one space: 440

Thus, the sample proportion is: $\hat{p}=\dfrac{440}{525}\approx0.838$

The z-score for a 90% confidence interval is: 1.645

The formula for determining the confidence interval:

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.838\pm (1.645)\sqrt{\dfrac{0.838(1-0.838)}{525}}\\\\=0.838\pm (1.645)\sqrt{0.00025858285}\\\\=0.838\pm (1.645)(0.01608)\\\\= 0.838\pm0.0265\\\\=(0.838-0.0265,\ 0.838+0.0265)\\\\=(0.8115,\ 0.8645)$

Consequently, a 90% confidence interval for the proportion of people who leave one space after a period is: (0.8115, 0.8645)

3 0

3 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

3 months ago

Ms. Gallegos burns 236 calories riding her bike each hour. She wants to burn more than 590 calories riding her bike at the same

Svet_ta [12734]

Response:

354

Detailed explanation:

5 0

3 months ago

Triangle X Y Z is shown. The length of side X Y is (9 n + 12) feet and the length of side X Z is (15 n minus 6) feet. Angles X Y

babunello [11817]

Answer:

$n = 3$

$XY = 39$

$XZ = 39$

Step-by-step explanation:

Given

$XY = 9n + 12$

$XZ = 15n - 6$

Solving for (a): n

To find n;

We will utilize

$XY = XZ$

Substituting values for XY and XZ

$9n + 12 = 15n - 6$

$9n - 15n = -12 - 6$

$-6n = -18$

Dividing by -6

$n = 3$

Solving for (b) XY

$XY = 9n + 12$

Substituting 3 for n

$XY = 9 * 3 + 12$

$XY = 27 + 12$

$XY = 39$

Solving for (c): XZ

$XZ = XY$

$XZ = 39$

7 0

3 months ago