50 (1/3) mph
47 (1/2) = 47.5 mph
47.5 / 50 (1/3) =
<span>
<span>
<span>
0.943708609271523
</span>
</span>
</span>
hours
times 60 =
<span>
<span>
<span>
56.6225165563
</span>
</span>
</span>
minutes =
56 minutes <span><span><span>37.3509933775
</span>
seconds
</span></span>
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).
K=5
if you add 5 to f(x) it will move up five and become g(x)
Answer:

Step-by-step explanation:

Rewrite the expression

Add

Therefore,
is equal to
.
I hope this helps!