<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>
6/7 is in fraction format . . . for it to be in decimal format it would be 0.8571 . . .
Answer:
0.258819045...
Step-by-step explanation:
just plug it into a calculator
Answer:
1.5 kg
Step-by-step explanation:
17*92 g = 1564
1564/1000= 1.564 kg
Answer:2.4
Step-by-step explanation:
12 * 20% = 12 * 0.20 = 2.4 hours
It will not last the entire shift.
