A(-7,-4) B(-2,0)
√[(x'-x)^2+(y'-y)^2]
√(-2-(-7)^2+(0-(-4)^2
√(5^2)+(4^2)
√25+16
√41
the distance is approximately 6.4 units
Because most of the area under any normal curve falls within a limited range of the number line
Answer:
the answer is 60%
Step-by-step explanation:
36/60=3/5
3/5= 0.6
0.6 x 100=60%
