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
Answer:
(x²+1)(x-3)
Step-by-step explanation:
Factorize x³ + x - 3x² -3
x³-3x²+x-3
Factorize
x²(x-3)+1(x-3)
= (x²+1)(x-3)
Hence the factorized form is (x²+1)(x-3)
Answer:
1305 female students
Step-by-step explanation:
Total no. of students = 2900
Percentage of students male = 55%
No. of students male = 55% of 2900
= 55/100*2900
= 1595
No. of female students = 2900 - 1595 = 1305
Answer:
2πr=72
=2*22/7*r=72
=r=72*7/44=504/44=252/22=126/11=11.45
=πr theta /180=22/7*11.45*60/180
=11.99=12(approx)
answer is (a) 12