The class of 2004 holds a reunion each year. In 2005, 87.5% of the 120 graduates attended. In 2006, 90 students went, and in 2007, 75 students went. About how many students do you predict went to the reunion in 2010

15 students

Step-by-step explanation:

% of students attended reunion in 2005 = 87.5% of 120

no of students attended reunion in 2005 = 87.5/100  * 120 = 105.

no of students attended reunion in 2006 = 90

no of students attended reunion in 2007 = 75

So , the series for the given year is

105, 90, 75

we can see that this arithmetical progression series

in  arithmetical progression series

common difference is = nth term - (n-1)th term

lets takes 3rd and second term as  nth term and (n-1)th term

common difference(d) is =75 - 90 = -15

n th term for any arithmetical progression series is given by

nth term = a + (n-1) d

where a is the first term

n is the nth term

d is the common difference

we have to calculate number of students in reunion of of 2010

as reunion began from 2005

2005,2006,2007,2008,2009,2010

so, 2010 is 6th term in series of 105, 90, 75

6th term =  a + (n-1) d

a is 90

6th term =  90+ (6-1)* -15

6th term =  90+ (5)* -15 = 90 - 75 = 15.

Thus, 15 students will attend reunion of 2010

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value