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1 answer:
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Answer:
The correct option is (b).
Step-by-step explanation:
If X N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
Answers:
f(-3) = -1
f(-1) = 2
f(3) = 5
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Explanations:
Draw a vertical line through -3 on the x axis. Make sure this vertical line crosses the orange graph. Mark this point of intersection and then draw a horizontal line from this point to the y axis. You should find that it touches -1 on the y axis. Check out the attached image to see this process in action.
Those steps basically say that x = -3 leads to y = -1. Since y = f(x), this means f(-3) = -1
Through similar steps, you should find that f(-1) = 2 and f(3) = 5
Note that the open hole/circle does not count. So if you land on an open hole, just keep going until you hit a closed circle on the other piece of the graph.
Answer:
- QN = 3√7
- OI = √7
Step-by-step explanation:
The ratio of corresponding sides DN and KI is 12 : 4 = 3 : 1. The same ratio applies to altitudes DQ and KO. Since the difference between these altitudes is 6 and the difference between their ratio units is 3-1 = 2, each ratio unit must stand for 6/2 = 3 units of linear measure. That is, ...
DQ = (3 units)·3 = 9 units
KO = (3 units)·1 = 3 units
Then the base lengths QN and OI can be found from the Pythagorean theorem:
KI² = KO² +OI²
4² = 3² +OI²
OI = √(16 -9)
OI = √7
QN = 3·OI = 3√7
