Solve for k, the constant of variation, in a joint variation problem where y = 520, x = 5, and z = 6.5
1 answer:
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In the expression given, the "n" is referring to a certain term, in this case, it would be the 16th term. So in order to find the 16th term, all you have to do is substitute n as 16 in the expression, so the answer is 
. So the 16th term of the sequence is -29.
Answer:
The probability of a random unit is deemed faulty if it scores below 320.
P( X < 320 ) = 0.0505
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given mean of the Population 'μ' = 374
Variance of the Population σ² = 1090
Standard deviation of the Population 'σ ' = √1090 = 33.01
Given 'X' be a random variable in normal distribution
Z = - 1.64
The probability of a random unit is deemed faulty if it scores below 320.
P( X < 320 ) = P( Z < - 1.64 )
= 1- p(Z > 1.64)
= 1 - ( 0.5 + A(1.64)
= 0.5 - 0.4495
= 0.0505
The probability of a random unit is deemed faulty if it scores below 320.
P( X < 320 ) = 0.0505
let's solve :
- 2 [30 - (1 + 4)²]
- 2 [30 - (5)²]
- 2 [30 - 25]
- 2 [5]
- 10