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MrRa
2 months ago
11

If the letters in the word poker are rearranged, what is the probability that the word will begin with the letter k and end with

the letter p?
Mathematics
1 answer:
Inessa [12.5K]2 months ago
3 0
The likelihood stands at 6 out of 120, or 1/20, equivalent to a 5% probability. You can calculate this by defining the total number of outcomes. When organizing items within a set, the factorial function on your calculator, denoted by "!", is a straightforward way to arrive at a solution. This function indicates the product of all integers less than that number down to 1 multiplied together. Therefore, since there are 5 letters in the word poker, we utilize 5! which can be expanded to 1*2*3*4*5 = 120. Next, we must identify how many combinations begin with k and conclude with p. There are precisely 6 such configurations: koerp, korep, kreop, kroep, kerop, and keorp. Finally, we divide the number of favorable outcomes by the total possibilities.
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Finally, suppose m1→∞, while m2 remains finite. what value does the the magnitude of the tension approach?
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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:
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Rearranging:
m₁*g - T = (m₁*T)/m₂ - m₁*g
2*m₁*g = T * (1 + m₁/m₂)
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3 months ago
Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a per
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Response: (0.8115, 0.8645)

Step-by-step outline:

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

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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)

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