The depth of the water at the end of a pier changes periodically along with the movement of tides. On a particular day, low tide

The equivalent ratio is 3/4.

The tension does not approach infinity.
<span>Let's analyze free body diagrams (FBDs) for each mass, considering the direction of motion of m₁ as positive.

For m₁: m₁*g - T = m₁*a

For m₂: T - m₂*g = m₂*a

Assuming a massless cord and pulley without friction, the accelerations are the same.

From the second equation: a = (T - m₂*g) / m₂

Substitute into the first:
m₁*g - T = m₁ * [(T - m₂*g) / m₂]
Rearranging:
m₁*g - T = (m₁*T)/m₂ - m₁*g
2*m₁*g = T * (1 + m₁/m₂)
2*m₁*m₂*g = T * (m₂ + m₁)
T = (2*m₁*m₂*g) / (m₂ + m₁)
Taking the limit as m₁ approaches infinity:
T = 2*m₂*g

This aligns with intuition since the greatest acceleration m₁ can have is -g. The cord then accelerates m₂ upward at g while gravity acts downward, leading to a maximum upward acceleration of 2*g for m₁.</span>

The final amount comes to $2313.51. Explanation: We compute the future value of each cash flow and aggregate them. Initially, $700 is deposited after year one. Considering a timeframe of three years at an interest rate of 6%. Next, $500 is deposited at the end of the second year, maturing in two years. Finally, $300 is deposited after three years, maturing in one year. Moreover, an additional $600 is deposited at the end of year four with no interest accrued on that amount. Therefore, the terminal value equals $833.71 plus $561.80 plus $318 plus $600 totals $2313.51.

Answer:

160/1001, 175/1001

Step-by-step explanation:

i) We calculate:

₈C₁ methods to select 1 new camera from a selection of 8

₆C₃ methods to select 3 refurbished cameras from a selection of 8

₁₄C₄ methods to select 4 cameras from the total of 14 cameras

The probability formula is:

P = ₈C₁ ₆C₃ / ₁₄C₄

P = 8×20 / 1001

P = 160 / 1001

P ≈ 0.160

ii) For at most one new camera, it means we want either one new camera or none at all. We've calculated the probability of selecting one new camera already. The probability of not selecting any new camera is equivalent to selecting 4 refurbished cameras:

P = ₆C₄ / ₁₄C₄

P = 15 / 1001

Therefore, the combined probability is:

P = 160/1001 + 15/1001

P = 175/1001

P ≈ 0.175

Respuesta:

P = 2/7 = 0.2857 o 28.57%

Explicación paso a paso:

En primer lugar, sabemos que hay 15 bolas y necesitamos identificar cuáles son pares y superiores a 10.

Por lo tanto, debemos calcular inicialmente la probabilidad de que el número obtenido sea par.

Los números pares son 2, 4, 6, 8, 10, 12, 14

Contamos 7 números de 15 ---> P(B) = 7/15

De esos números, solo dos son mayores a 10, que son 12 y 14, así que: P(A|B) = 2/15

Para encontrar la probabilidad de obtener un número par mayor que 10:

P(A/B) = P(A|B) / P(B)

P(A/B) = 2/15 / 7/15 = 2/7 = 0.2857

Para calcular el porcentaje: 0.2857 * 100 = 28.57%.