This function has a straight line graph and thus has no limit on the range.
There are 6 "two-month periods" in a year, where the tree will grow 14 inches each period. Thus, the tree grows a total of:
![6 \cdot 14\, \textrm{in} = \boxed{84 \,\textrm{in}}](https://tex.z-dn.net/?f=%206%20%5Ccdot%2014%5C%2C%20%5Ctextrm%7Bin%7D%20%3D%20%5Cboxed%7B84%20%5C%2C%5Ctextrm%7Bin%7D%7D%20)
The answer is B, 84 in.
Answer:
13 and 27
Step-by-step explanation:
x+y=40
x- smaller
2y-x=41
x=2y-41
Now to plug it into our first equation:
3y-41=40
3y=81
y=27
x=13
Answer:
y=-6x+33
Step-by-step explanation:
Perpendicular lines meet the following condition:
m1*m2=-1
From the first line equation, we obtain m1 if we write the given equation in a proper manner. 6y=x-12 becomes y=x/6 -12,
m1=1/6 since is the coefficient of 'x' variable.
Now, to obtain 'm2' we use the condition for perpendicular lines
(1/6)*m2=-1
m2=-6
Thus, our new line have an equation like the following y=mx+b, where m=-6
Now, we need to eval the given point (6,-3) in the equation in from above to obtain the value of 'b'
-3= -6*(6) + b
Leading us to, b=33
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:
![Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B48-60%7D%7B12%7D%3D%5Cfrac%7B-12%7D%7B12%7D%3D-1.0)
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).