1. Determine the area under the standard normal curve to the left of z=2.10.
1 answer:
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4(q+3) + 3q = -30
4q +12 + 3q = -30
7q+12= -30
7q = -12-30
7q = -42
q = -6
p = -3
4q +12 + 3q = -30
7q+12= -30
7q = -12-30
7q = -42
q = -6
p = -3
Answer:
Step-by-step explanation:
a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.
b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).
c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following
where