We are looking to find P(X>60 students)
X is normally distributed with mean 50 and standard deviation 5
We need to find the z-score of 60 students
To find the probability of P(Z>2), we can do 1 - P(Z<2)
So we read the probability when Z<2 which is 0.9772, then subtract from one we get 0.0228
The number of students that has score more than 60 is 0.0228 x 1000 = 228 students
Answer:
the y intercept would be at (0,-6)
Answer:
(-2,-7) answer (-1,14)
3 questions answer (4,-11)
(5,10) answer ( 10,-15)
Answer:
Step-by-step explanation:
Set up synthetic division, starting with x - 1 as the divisor, and then (from that) 1:
1 / 1 0 0 -1 The coefficients of x^2 and x are zero here.
1 1 1
--------------------------
1 1 1 0
Since the remainder is zero, 1 is a root and x - 1 is a factor of the original
x^3 - 1 = 0.
Use 3.14 multiplied by the base and width