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Area of ABC : AB*AC/2
the maximum of the parabola is reached at x=-4/(2*(-1))=2 hence A is at (2,0) and B is at (2,(-2)^2+4*2+C)=(2,12+C)
C is the second root (x-intersect), which we can find :
determinant1 : D=16-4*(-1)*C=4(4+C) thus the second root is at x=
Hence the area of the triangle is
hence 
.
We remark that
Hence 4+C=16 thus C=12
the maximum of the parabola is reached at x=-4/(2*(-1))=2 hence A is at (2,0) and B is at (2,(-2)^2+4*2+C)=(2,12+C)
C is the second root (x-intersect), which we can find :
determinant1 : D=16-4*(-1)*C=4(4+C) thus the second root is at x=
Hence the area of the triangle is
We remark that
Hence 4+C=16 thus C=12
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Answer:
Required Probability = 0.97062
Step-by-step explanation:
We are given that the weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4016 grams and a standard deviation of 532 grams.
Let X = weight of the newborn baby, so X ~ N()
The standard normal z distribution is given by;
Z = ~ N(0,1)
Now, probability that the weight will be less than 5026 grams = P(X < 5026)
P(X < 5026) = P( <
) = P(Z < 1.89) = 0.97062
Therefore, the probability that the weight will be less than 5026 grams is 0.97062 .