Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
<span>First, the inequality needs to be solved. The first step is to subtract 8 from both sides of the inequality, leading to 5r < 55. Dividing 5 out from both sides, this will leave r < 11. Next, to form a set notation, the inequality is written in such form: {r | r < 11}.</span>
<u>Answer:</u>
<u>Step-by-step explanation:</u>
We know from the question that the student earned $12.50 <em>per hour</em>.
Using this information, we can say that if the student worked for <em>h </em>hours, they would make a total of 12.50 × <em>h </em>dollars.
We also know that the total money they earned is $2500.75.
∴ Therefore, we can set up the following equation:
From here, if we want to, we can find the number of hours worked by simply making <em>h</em> the subject of the equation and evaluating:
<em>h </em>=<em> </em>
= 200.6 hours
8 am at 0
8:18 at 4.5
8:48 at 7.5 miles ???
18 min = .3 hour
48 min = .8 hour
in first .3 hours speed = 4.5/.3 = 15 mph
in next .5 hours = speed = 3/.5 = 6 mph
average speed for the whole trip = 7.5 miles/.8 hours = 9.375 mph
