Solution:
To test the hypothesis is that the mean ozone level is different from 4.40 parts per million at 1% of significance level.
The null hypothesis and the alternative hypothesis is :
The z-test statistics is :
z = -1.37
The z critical value for the two tailed test at 99% confidence level is from the standard normal table, he z critical value for a two tailed at 99% confidence is 2.57
So the z critical value for a two tailed test at 99% confidence is ± 2.57
Conclusion :
The z values corresponding to the sample statistics falls in the critical region, so the null hypothesis is to be rejected at 1% level of significance. There is a sufficient evidence to indicate that the mean ozone level is different from 4.4 parts per million. The result is statistically significant.
Your answer is 8.4 X 10^7.
Hope this helped!
Answer:
y = 56
Step-by-step explanation:
the midsegment SV is half the length of the side TU , that is
y - 9 = 
(y + 38) ← multiply both sides by 2 to clear the fraction
2y - 18 = y + 38 ( subtract y from both sides )
y - 18 = 38 ( add 18 to both sides )
y = 56
Solution for F(x)=6x^2-900 equation:
<span>Simplifying
F(x) = 6x2 + -900
Multiply F * x
xF = 6x2 + -900
Reorder the terms:
xF = -900 + 6x2
Solving
xF = -900 + 6x2
Solving for variable 'x'.
Reorder the terms:
900 + xF + -6x2 = -900 + 6x2 + 900 + -6x2
Reorder the terms:
900 + xF + -6x2 = -900 + 900 + 6x2 + -6x2
Combine like terms: -900 + 900 = 0
900 + xF + -6x2 = 0 + 6x2 + -6x2
900 + xF + -6x2 = 6x2 + -6x2
Combine like terms: 6x2 + -6x2 = 0
900 + xF + -6x2 = 0 </span>
Y=x-3 it's one of many equations
