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evablogger [386]
3 years ago
12

My life is hellpless.... could you please help? i cant even understand this I'm only 11.

Mathematics
2 answers:
andrezito [222]3 years ago
7 0

Answer:

<em>Y-intercept:</em> -4

<em>Slope:</em> 2

<em>Equation:</em> y = 2x - 4

Digiron [165]3 years ago
4 0

Answer:

y-intercept: -4

slope: 2

equation: y = -4x + 2

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Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar
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P(A) = \frac{1}{9}

P(B) = \frac{1}{3}

P(C) = \frac{1}{729}

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Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

P(M) = \frac{1}{3}.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is P(A) = \frac{1}{9}. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}.

C. If the probability that two people have been hired on a specific day of the week is 3(\frac{1}{3}) ^ 2, then the probability that 7 people have been hired on the same day is:

P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}

D. The probability is \frac{1}{729}. This number is quite close to zero. Therefore it is an unlikely bastate event.

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