The general formula: y=mx+b
(6,2) ⇒ x=6,y=2
m=3
b=?
2=3*6+b
2=18+b
b=-16
<u>y=3m-16</u>
        
             
        
        
        
Answer:
I have made it in above figure 
 
        
                    
             
        
        
        
Answer:
The correct option is (c). 
Step-by-step explanation:
The complete question is:
The data for the student enrollment at a college in Southern California is:
                     Traditional          Accelerated            Total
                   Math-pathway     Math-pathway
Female              1244                       116                   1360
Male                  1054                       54                    1108
Total                  2298                     170                  2468
We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the student is female. Which of the following pairs of probabilities is not a useful comparison?
a. 1360/2468 and 116/170
b. 170/2468 and 116/1360
c. 1360/2468 and 170/2468
Solution:
If two events <em>A</em> and <em>B</em> are independent then:

In this case we need to determine whether a student enrolled in an accelerated math pathway is independent of the student being a female.
Consider the following probabilities:

If the two events are independent then:
P (F|A) = P(F)
&
P (A|F) = P (A)
But what would not be a valid comparison is:
P (A) = P(F)
Thus, the correct option is (c). 
 
        
             
        
        
        
A² = b² + c² - 2bc cosA
a² = 10² + 14² - 2*10*14 cos54
a² = 100 + 196 - 280 * cos54
a = 
 
 a = 11.46