Answer:
In kJ/kg.K - 1.005 kJ/kg degrees Kalvin.
In J/g.°C - 1.005 J/g °C
In kcal/ kg °C 0.240 kcal/kg °C
In Btu/lbm-°F 0.240 Btu/lbm degree F
Step-by-step explanation:
given data:
specific heat of air = 1.005 kJ/kg °C
In kJ/kg.K
1.005 kJ./kg °C = 1.005 kJ/kg degrees Kelvin.
In J/g.°C
In kcal/ kg °C
For kJ/kg. °C to Btu/lbm-°F
Need to convert by taking following conversion ,From kJ to Btu, from kg to lbm and from degrees C to F.
1.005 kJ/kg C = 0.240 Btu/lbm degree F
Answer:
Z-score 1.96
Margin of error d= 0.04323
Step-by-step explanation:
Hello!
The study variable in this case is
X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.
n= 500
The parameter of interest is the proportion of people that opposes it.
The estimated proportion is p'= 300/500= 0.6
To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:
p'≈N(p;p(1-p)1/n)
The mean of this distribution is p and the variance is p*(1-p)*1/n
To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:
p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)
The formula for the confidence interval is:
p' ± *
Where " *
" represents the margin of error of the interval.
Now for a confidence level of 0.95 the value of Z is
The estimated standard error is already calculated:
The margin of error (d) is then:
d= *
= 1.965 * 0.022
d= 0.04323
I hope it helps!