I’m sorry I don’t get what your asking?
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.
100 is ten times as much as 1/10 of a hundred
ur multiplying n then dividing by ten so they cancel each other out
theres no table so idk if this is what u wanted but that's the answer
hope it helps
Answer: The LCM of 8, 9, and 19 is 1,368.
Prime factorization of the numbers:
8 = 2 x 2 x 2
9 = 3 x 3
19 = 19
2 x 2 x 2 x 3 x 3 x 19 = 1,368.
Hope this helps! Have a great day!
Answer:
hope this helps!
Step-by-step explanation:
From y(t) = −16t^2 − 3t + 300 find at what time the fertilizer hits the ground (has height 0)
0 = −16t^2 − 3t + 300
Using the quadratic formula (minus b plus or minus the square root of b squared minus 2 a c over 2 a) you get
t = 4.237 seconds
So the horizontal distance is 85t = 85 * 4.237 = 360.145 feet