Answer:
P'(-9, 3)
Step-by-step explanation:
The transformation for rotation counterclockwise about the origin by some angle α will be ...
(x, y) ⇒ (x·cos(α) -y·sin(α), x·sin(α) +y·cos(α))
When the angle is α = 90°, this reduces to ...
(x, y) ⇒ (-y, x)
You have (x, y) = (3, 9), so the image point is P'(-9, 3).
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<em>Additional comment</em>
The "work" is applying the right sign and choosing the right coordinate to fill in the values (-y, x). I find it easier to look up the transformations for ±90° and 180°, rather than try to derive them from first principles every time. Your text may have a list:
- 90° CCW or 270° CW: (x, y) ⇒ (-y, x)
- 90° CW or 270° CCW: (x, y) ⇒ (y, -x)
- 180°: (x, y) ⇒ (-x, -y)
Answer:
12
Step-by-step explanation:
Answer:
1/3 · -x < 12
Step-by-step explanation:
one third meaning 1/3
"of" meaning multiplication
"opposite" of a "number" meaning negative x
Less than 12 meaning > 12
I believe the answer to this is "If they babysit for 5 hours, Jayna will earn more" because Hannah charages 7$ an hour and Jayna charges 10$ an hour so Jayna will have more money than Hannah in five hours.
Answer:
The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What lifetime should the company promote for these bulbs, whereby only 2% burnout before the claimed lifetime?
This is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.
The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
