Sammi has $125.75 in her savings account. She deposits $25.50 into the account each month for the next 6 months. How much is in

The probability that Jane will go to a ballgame (event A) on a Monday is 0.73, and the probability that Kate will go to a ballga

Inessa [12570]

Events A and B are termed independent when

$Pr(A\cap B)=Pr(A)\cdot Pr(B),$

if not, events A and B are classified as dependent.

The events A, B and A∩B are:

A - Jane plans to attend a ballgame on Monday;
B - Kate intends to go to a ballgame on Monday;
A∩B - Both Kate and Jane will be at the ballgame on Monday.

$Pr(A)=0.73,\ Pr(B)=0.61,\ Pr(A\cap B)=0.52.\\ \\ Pr(A)\cdot Pr(B)=0.73\cdot 0.61=0.4453\neq 0.52=Pr(A\cap B).$

Answer: events A and B are dependent

4 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

(c) 1.89x10⁻⁴

(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

16.A guidance counselor is planning schedules for 30 students. Sixteen students say they want to take French, 16 want to take Sp

PIT_PIT [12445]

Response: 7

Detailed explanation:

A Venn diagram can help visualize this problem.

There are a total of 5 students interested in both French and Latin.

Out of these, 3 students also want to learn Spanish, meaning only 2 students want solely French and Latin.

Moreover, there are 5 students who wish to study only Latin.

This results in 1 student wanting both Latin and Spanish, calculated by 11 - 5 - 3 - 2.

There are 8 students who desire only Spanish, and 4 students who want both Spanish and French.

In the same manner, those wishing to study only French amount to 16 - 4 - 3 - 2 = 7.

8 0

3 months ago

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The pH of a person’s eyes is 7.2. Therefore, the ideal pH for the water in a swimming pool is between 7.0 and 7.6. Write a compo

lawyer [12517]

Answer:

7.0 < x < 7.6

X represents the optimal pH level for the swimming pool water

5 0

2 months ago