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blsea [12.9K]
3 years ago
14

What approximate error rate should the manager expect for a worker who receives 6 hours of training?

Mathematics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

c.0.13

Step-by-step explanation:

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Which expression is not equivalent to
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B i think I'm kinda sure tho

b

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The cube root of rvaries inversely with the square of s. Which two equations model this relationship
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Answer:

3√r = k/s²

s²r^1/3 = k

Step-by-step explanation:

Cube root of r = 3√r

Square of s = s²

cube root of r varies inversely with the square of s

3√r = k/s²

Cross product

3√r * s² = k

s²r^1/3 = k

Note:

r^1/3

= 3√r

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Long addition for integers
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-9 + 6 + (-7) +3=
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Read 2 more answers
A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time).
likoan [24]

Answer:

(a). 72.9%.

(b). 13.6 hr.

Step-by-step explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this question/problem;

=> "A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time). "

=> The worker produces 10 welded assemblies on the second day, and the 10th assembly on the second day takes 30 minutes"

So, we will be making use of the Crawford learning curve model.

T(7) + 10 = T (17) = 30 min.

T(7) = T1(7)^b = 45.

T(17 ) = T1(17)^b = 30.

(T1) = 45/7^b = 30/17^b.

45/30 = 7^b/17^b = (7/17)^b.

1.5 = (0.41177)^b.

ln 1.5 = b ln 0.41177.

0.40547 = -0.8873 b.

b = - 0.45696.

=> 2^ -0.45696 = 0.7285.

= 72.9%.

(b). T1= 45/7^ - 045696 = 109.5 hr.

V(TT)(17) = 109.5 {(17.51^ - 0.45696 – 0.51^ - 0.45696) / (1 - 0.45696)} .

V(TT) (17) = 109.5 {(4.7317 - 0.6863) / 0.54304} .

= 815.7 min .

= 13.595 hr.

8 0
3 years ago
Javier invests $2,500 into an account 3.5% simple interest . What will be the total amount in the account after 5 years if no ot
katovenus [111]

Answer:d

Step-by-step explanation:

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