Answer:
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
Step-by-step explanation:
We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.
We have to calculate a 95% confidence interval for the proportion.
The sample proportion is p=0.26.
The standard error of the proportion is:
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the population proportion is (0.192, 0.328).
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
The end of the ray stops the x values from proceeding left at x=0. So your domain is from that point on to infinity. In your solution set x >= 0, since the arrow continues on the right side where x's are positive.
Answer is 4 lb.,12 oz.
As one pound has 16 ounces.
When we convert the given 5 lb. into ounces we multiply it by 16.
5 lb = 5 x 16 oz
5 lb = 80 oz
When we subtract 4 oz from 80 oz
We get
80 - 4 = 76 oz
Which is
64 oz + 12 oz
(16 x 4) oz + 12 oz
Which will be
4 lb. + 12 oz
So the answer will be
4 lb., 12 oz.
The unknown side is 9. Because 9 x 5= 45, 45 being the perimeter, that is 5 times the unknown side. And 45 is the sum of 9, 16, and 20.
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
