Exercise 5.6.6: selecting a committee of senators. about a country has two political parties, the demonstrators and the repudiators. suppose that the national senate consists of 100 members, 44 of which are demonstrators and 56 of which are rupudiators. (a) how many ways are there to select a committee of 10 senate members with the same number of demonstrators and repudiators? suppose that each party must select a speaker and a vice speaker. how many ways are there for the two speakers and two vice speakers to be selected? feedback?

There are 4,148,350,734,528 ways

Step-by-step explanation:

We have

(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?

We want to choose 5 Demonstrators and 5 Repudiators. The number of ways to do this is and respectively. Therefore, the number of ways to select the committee is given by:

(b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

This is because there are (in the case of Demonstrators) 44 possibilities to choose an speaker and after choosing one, there would be 43 possibilities to choose a vice speaker. The same situation happens in the case of Repudiators.

.

i would have to say it is c.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.