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
69,010 is the total cost of ???
Answer:
Range is 44
mean is 29.7
variance is 289.56
standard deviation is 17.016
Step-by-step explanation:
The data given as follows
3, 24, 30, 47, 43, 7, 47,13, 44, 39
Range = maximum - minimum
= 47 -3 = 44
mean =
= 297/10 = 29.7
variance =
= 289.56
SD =
= 17.016
The standard deviation of the given data is not a good measure as the range and standard deviation are far apart. The standard deviation gives how the given data is dispersed from the mean or expected value, while the range gives a rough idea of variability of the data.
Add <span>6</span> to both sides
<span>2x-6+6=4+6</span>
Simplify <span>4<span>+6</span></span> to <span>10</span>
<span>2x=10</span>
Divide both sides by <span>2</span>
<span>2x/2=10/2</span>
Simplify 10/2
<span><span>x=<span>5</span></span><span>
</span></span>
Answer:
yh its true
Step-by-step explanation: