<span>The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50. what is the probability that a student uses more than 580 minutes?
Given
μ=500
σ=50
X=580
P(x<X)=Z((580-500)/50)=Z(1.6)=0.9452
=>
P(x>X)=1-P(x<X)=1-0.9452=0.0548=5.48%
</span>
Answer:
which principle prevents a brach from abusing its power
Step-by-step explanation:
Answer:
compostion
Step-by-step explanation:
comPOSTION
10/100=0.1
1/4=0.25
0.1*0.25=0.025
Answer:
The distance is 9.21954445... or just 9.22
