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Marysya12 [62]
3 years ago
14

In a test of corrosion resistance, a sample of 62 Incoloy steel specimens were immersed in acidified brine for 4 hours, after wh

ich each specimen had developed a number of corrosive pits. The maximum pit depth was measured for each specimen. The mean depth was 850 μm with a standard deviation of 153 μm. The specification is that the population mean depth μ is less than 900 μm.
Find the P-value for testing H0 : μ ≥ 900 versus H1 : μ < 900.

Mathematics
1 answer:
Dmitry [639]3 years ago
7 0

Answer:

p-value = 0.0063

Step-by-step explanation:

PLEASE CHECK ATTACHMENT FOR COMPLETE SOLUTION AND STEP BY STEP EXPLANATION

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andrew11 [14]

Answer:

2. B

3. B

4. 2(7k+5h)

5. $325

Step-by-step explanation:

2) "X decreased by the sum of 2x + 5" Decreased means to be lowered, so that means this problem has subtraction in it. It also says x is being decreased BY the sum of 2x + 5 so that means 2x + 5 needs to be solved, which calls for parentheses. B best fits the description.

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3 years ago
Solve the system of equations. y=-3x+4 -4x+y=-10​
Angelina_Jolie [31]

Answer:

Step-by-step explanation:

In this particular case we have the following system of equations:

y

=

−

3

x

+

4

[

E

q

.

1

]

x

+

4

y

=

−

6

[

E

q

.

2

]

Substituting

[

E

q

.

1

]

in

[

E

q

.

2

]

:

x

+

4

(

−

3

x

+

4

)

=

−

6

Applying the distributive property on the left side:

x

−

12

x

+

16

=

−

6

Simplifying

:

−

11

x

=

−

22

Solving for

y

:

x

=

−

22

−

11

=

2

Substituting

x

=

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in

[

E

q

.

1

]

:

y

=

−

3

(

2

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+

4

=

−

2

Therefore

, the solutions are

x

=

2

and

y

=

−

2

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3 years ago
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jeka94
2(5+y)= 18 than y= 4
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Answer:

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Mazyrski [523]

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3 years ago
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