Let the random variable X represent the weight of male black bears before they begin hibernation. Research has shown that X can be modeled by a Normal distribution with a mean of 250 pounds and a standard deviation of 50 pounds. What is the probability that a randomly selected male black bear will weigh more than 325 pounds before beginning hibernation? a
0.0668
b
0.7486
c
0.2514
d
0.8531
e
0.9332

Idon’t have want a lot more

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to