Find the area under the standard normal distribution curve to the left of z=-2.15 and to the right of z=1.62
1 answer:
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Answer:
Step-by-step explanation:
From the step given step 6
(b² — 4ac) / 4a² = (x + b/2a)²
Mistake in question it is (x + b/2a)²
Step 7 given
±√(b² —4ac) /2a = x + b/2a
Mistake in question it is over 2a and not 1a.
1. The step taken from step 6 to step 7 is taking square roots of both sides
(b² — 4ac) / 4a² = (x + b/2a)²
Taking square of both sodes
√(b²—4ac) / √4a² = √(x + b/2a)²
√(b²—4ac) / 2a = x + b/2a
This is the required step 7.
Now, subtracting b/2a from both sides
√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a
√(b²—4ac) / 2a — b/2a = x
(√(b²—4ac) — b)/2a = x
x = [—b ± √(b²—4ac)] / 2a
So, this is the required formula method
The discriminant is D = b²—4ac.
For this case we must find an expression equivalent to:
According to the unit circle, is in the third quadrant and equals 210 degrees.
Then, we must find an expression equivalent to:
We have to:
It is in the fourth quadrant and is equivalent to 330 degrees, according to the unit circle.
Answer:
Option D