<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 is b.Step-by-step explanation:
Answer: 
<u>Step-by-step explanation:</u>
Average rate of change is the slope (m) between the two coordinates (-1, -1) and (1, -2).



Answer:
35-5a-3a-12-4=-8a+19
4a-12+7=4a-5
Unequal
Step-by-step explanation: