One function you would be trying to minimize is
<span>f(x, y, z) = d² = (x - 4)² + y² + (z + 5)² </span>
<span>Your values for x, y, z, and λ would be correct, but </span>
<span>d² = (20/3 - 4)² + (8/3)² + (-7/3 + 5)² </span>
<span>d² = (8/3)² + (8/3)² + (8/3)² </span>
<span>d² = 64/3 </span>
<span>d = 8/sqrt(3) = 8sqrt(3)/3</span>
Answer: 0.8461
Step-by-step explanation:
Given : 
Let x be the random variable that represents the cost for the hospital emergency room visit.
We assume that cost for the hospital emergency room visit is normally distributed .
z-score for x=1000 ,
Using z-value table , we have
P-value =P(x>1000)=P(z>-1.02)=1-P(z≤ -1.02)=1-0.1538642
=0.8461358≈0.8461 [Rounded nearest 4 decimal places]
Hence, the probability that the cost will be more than $1000 = 0.8461
Answer:
9in
Step-by-step explanation:
54 = 1/2(12*x)
54 = 6*x
x = 9
Check
12 x 9 = 108
108/2 = 54
-6+10c
I think, reorder the terms: -6+9c+c= -6+10c
<h3>E
xplanation:</h3>
Replace cos^2(θ) with 1-sin^2(θ), and cot(θ) with cos(θ)/sin(θ).
cos^2(θ)cot^2(θ) = cot^2(θ) - cos^2(θ)
(1 -sin^2(θ))cot^2(θ) = . . . . . replace cos^2 with 1-sin^2
cot^2(θ) -sin^2(θ)·cos^2(θ)/sin^2(θ) = . . . . . replace cot with cos/sin
cot^2(θ) -cos^2(θ) = cot^2(θ) -cos^2(θ) . . . as desired