<span>With the mean of 40k and standard deviation of 5k, we need to find P(x>=30k). P((30k-40k)/5k) = P(z<-2) = 1-0.0228 which is equal to 0.9772. There is a 97.72% chances that the starting salary will be at least 30k.</span>
71 is the answer. If you round it, it would be 70.
1 Degree Radian = π (pie)/ 180 degrees
6 idk it’s making me answer
