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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
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:
<h2>By angle sum property:</h2>BAC+ABC+BCA=180°
90°+45°+x=180°
135°+x=180°
x=45°=BCA
<h2>since BCA=ABC=<u> </u>45° each </h2><h2>therefore,</h2>