-2(-5x - 10) + 30 = -110
2 answers:
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Answer:
- <u><em>Option b. just below 30%</em></u>
<u><em></em></u>
Explanation:
Please, see attached the <em>histogram that represents the distribution of acceptance rates (percent accepted) among 25 business schools in 2004. </em>
<em />
The<em> median</em> is the value that separates the lower 50% from the upper 50% of the data.
Since there are 25 business schools, the middle value is the number 13.
The height of each bar is the<em> frequency</em> or number of business school for that acceptace rate:
- The first bar has frequency of 1 school
- The second bar has frequency of 3 schools: cummulative frequency: 1+3=4.
- The third bar has frequency 5 schools: cummulative frequency 4 + 5 = 9.
- The fourth bar has frequency 3 schools: cummulative frequency: 9+3=12.
Then, the 13th value is on the next bar, the fifth bar.
The fifth bar has acceptance rates 25 ≤ rate < 30.
That means that the median acceptance rate is greater than or equal to 25 and less than 30.
Thus, the choice is the option <em>b. just below 30%.</em>
(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:
(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become:
β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:
(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become:
β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.