Start by reviewing your knowledge of natural logarithms. If we take the ln of both sides we get e^z=ln(1). Do the same thing again and wheel about the ln(ln(1)). There's going to be complex solutions, Wolfram Alpah gets them but let me know if you figure out how to do it?
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity
Sample 1 (aqueous film forming foam)
n₁= 5
X[bar]₁= 4.7
S₁= 0.6
Sample 2 (alcohol-type concentrates )
n₂= 5
X[bar]₂= 6.8
S₂= 0.8
Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?
The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:
t= ![\frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }](https://tex.z-dn.net/?f=%5Cfrac%7B%28X%5Bbar%5D_1%20-%20X%5Bbar%5D_2%29%20-%20%28mu_1%20-%20mu_2%29%7D%7BSa%2A%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%20%7D%20%7D%20%7D)
a) 95% CI
(X[bar]_1 - X[bar]_2) ±
*
Sa²=
=
= 0.5
Sa= 0.707ç

(4.7-6.9) ± 2.306* 
[-4.78; 0.38]
With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.
b.
The hypothesis is:
H₀: μ₁ - μ₂= 0
H₁: μ₁ - μ₂≠ 0
α: 0.05
The interval contains the cero, so the decision is to reject the null hypothesis.
<u>Complete question</u>
a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.
b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?
Answer:
64.8% of elephants were killed for ivory
Step-by-step explanation:
divide the number of elephants killed for ivory by the total number of elephants killed: 186÷287= 0.6780
multiply your answer by 100 to get a percentage: 0.6780 x 100= 67.8%
Answer:
The answer is 33.75
Step-by-step explanation:
You simply need to multiply 67.5 by .5 in order to find half of it.