Which solution is possible: 7+x <19 x=4 x=12 x=13 x=19 Other:

Mathematics

2 answers:

ZanzabumX [31]3 years ago

8 0

Answer:

none of the above, you need a lesser number

Tamiku [17]3 years ago

4 0

Answer:

x = 4

Step-by-step explanation:

Note that it is only less than, not less than or equal too. Numbers that are 11 & less are the only ones that work.

x = 4 is your only answer choice that is less than 11, therefore is your only answer.

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2 years ago

Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar

horsena [70]

Answer:

$P(A) = \frac{1}{9}$

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

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

Step-by-step explanation:

We know that:

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.

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

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: