The fourth choice is the correct one.
Answer: the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Step-by-step explanation:
Given data;
lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min
hence, height = 1 / ( 52.0 - 50.0) = 1 / 2
now the probability that the class length is between 50.8 and 51 min = ?
P( 50.8 < X < 51 ) = base × height
= ( 51 - 50.8) × 1/2
= 0.2 × 0.5
= 0.1 ≈ 10%
therefore the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Answer:
2.3%
Step-by-step explanation:
18000 = 15000( 1 + i)^8
Divide both sides by 15000
1.2 = (1 + i)^8
Take both sides to the 1/8 power
1.2^(1/8) = 1 + i
1.023051875220463 = 1 + i
Subtract 1 from both sides
i = 0.023051875220463
~ 2.3%
The maximum value of the objective function is 26 and the minimum is -10
<h3>How to determine the maximum and the minimum values?</h3>
The objective function is given as:
z=−3x+5y
The constraints are
x+y≥−2
3x−y≤2
x−y≥−4
Start by plotting the constraints on a graph (see attachment)
From the attached graph, the vertices of the feasible region are
(3, 7), (0, -2), (-3, 1)
Substitute these values in the objective function
So, we have
z= −3 * 3 + 5 * 7 = 26
z= −3 * 0 + 5 * -2 = -10
z= −3 * -3 + 5 * 1 =14
Using the above values, we have:
The maximum value of the objective function is 26 and the minimum is -10
Read more about linear programming at:
brainly.com/question/15417573
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Answer:
30 degrees
Step-by-step explanation:
u dot v=1*4+1*4+1*4+0*4=4+4+4+0=12
|u|=sqrt(1^2+1^2+1^2+0^2)=sqrt(3)
|v|=sqrt(4^2+4^2+4^2+4^2)=sqrt(4*4^2)=2*4=8
cos(theta)=u dot v/(|u||v|)
cos(theta)=12/(sqrt(3)*8)
cos(theta)=3/(sqrt(3)*2)
cos(theta)=sqrt(3)/2
theta=30 degrees