<span>Best Answer: 1) Q is positive therefore the process is endothermic. .769kj+-.810kj=-.041kj
2) Heat is released meaning it is exothermic. -66.9kj+45=-21.9kj
3) 7.29kj. Endothermic.</span>
Answer:
For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30
Step-by-step explanation:
For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30 i.e n ≥ 30 this is because the shape of a sample distribution depends on the sample size. But for normal distribution population, the sample means would be approximately a normal distribution even if the sample size is less than 30;
Divide the change by the initial number to get a fraction and multiply by 100% to get a percent. The percent you get will describe the change in area in the mathematical language of numbers.
Answer:
We accept H₀ with the information we have, we can say level of ozone is under the major limit
Step-by-step explanation:
Normal Distribution
population mean = μ₀ = 7.5 ppm
Sample size n = 16 df = n - 1 df = 15
Sample mean = μ = 7.8 ppm
Sample standard deviation = s = 0.8
We want to find out if ozono level, is above normal level that is bigger than 7.5
1.- Hypothesis Test
null hypothesis H₀ μ₀ = 7.5
alternative hypothesis Hₐ μ₀ > 7.5
2.-Significance level α = 0.01 we will develop one tail-test (right)
then for df = 15 and α = 0,01 from t -student table we get
t(c) = 2.624
3.-Compute t(s)
t(s) = ( μ - μ₀ ) / s /√n ⇒ t(s) = ( 7.8 - 7.5 )*4/0.8
t(s) = 0.3*4/0.8
t(s) = 1.5
4.-Compare t(s) and t(c)
t(s) < t(c) 1.5 < 2.64
Then t(s) is inside the acceptance region. We accept H₀
