Answer:
As 2.551 < 3.84 therefore we reject H0.
As 44.803 > 4.46 so we accept null hypothesis.
Step-by-step explanation:
The answer is attached.
There are three judges so v1 = 3-1= 2 and v2 = (4*2)= 8
There are five gymnasts so v1 = 5-1= 4 and v2= (4*2)= 8
For alpha = 0.05 we find the value of F1 and F2 from the table.
Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer
Answer:
72
Step-by-step explanation:
there are 60 seconds in a minute so about 6 9 seconds in a minute. then there is 12 pages per 9 seconds and since there is 6 9 seconds then you can just multiply 6 by 9 to get the answer 72
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