Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) = 
= 
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4
Answer:
B, A, F, H
Step-by-step explanation:
Answer:
The bottom cutoff heights to be eligible for this experiment is 66.1 inches.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
Mean of 69.0 inches and a standard deviation of 2.8 inches.
This means that 
What is the bottom cutoff heights to be eligible for this experiment?
The bottom 15% are excluded, so the bottom cutoff is the 15th percentile, which is X when Z has a pvalue of 0.15. So X when Z = -1.037.
The bottom cutoff heights to be eligible for this experiment is 66.1 inches.
Answer:
6 7/8
Step-by-step explanation:
To evaluate 5 + 1 7/8 we combine the integers 5 and 1 and to this sum add the fraction 7/8:
5 + 1 + 7/8 = 6 7/8 (six and seven eighths)
This will be easier to write, and a lot easier to read, if we temporarily
use another symbol ... say, 'Q' ... to represent ' sin(2x) ' .
Here we go:
Original equation: Q² - 0.5 Q = 0
Factor the left side: Q (Q - 0.5) = 0
This equation is true if either factor is zero:
-- If Q=0, then sin(2x) = 0
2x = 0°, 180°, 360°
x = 0°, 90°, 180°
-- If (Q-0.5) = 0, then Q = 0.5
sin(2x) = 0.5
2x = 30°, 150°
x = 15°, 75°
The whole collection of solutions
between 0° and 360° :
x = 0°, 15°, 75°, 90°, 180° .
