B) D) and probably C) I’m not sure
Answer:
274878070200
Step-by-step explanation:
4 to the power 11 = 4194304
4 to the power -8 = 0.00001525878
4194304/0.00001525878 = 274878070200
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
W=l-8
a=9
a=lw
9=(l-8)l
9=l²-8l
0=l²-8l-9
factor
0=(l-9)(l+1)
set to zero
l-9=0
l=9
l+1=0
l=-1, false length can't be negative
l=9
the length is 9 units long
1. Multiply the hourly rate by the number of hours worked ( 22.5h) add that to the flat fee ( 22.5h +35) and that needs to be equal to or greater than what he expects to make:
A. 22.5h + 35 ≥ 215 , with a solution of h ≥ 8
2. Same method as number 1:
C. $3 + 1.80m ≤ $25
3. Each fraction would need 1 negative sign and have the same numerator and denominator, so the answer would be B.6/-7
