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?
Hi there
The formula you meant
V (t)=27500 (0.84)^t
V (10)=27,500×(0.84)^(10)
V (10)=4,809.8 Round your answer to get 4810 ...this isthe value after
10 years
the initial value of the car is
27500
Hope it helps
Answer:
2.92
Step-by-step explanation:
4 percent *73 =
(4:100)*73 =
(4*73):100 =
292:100 = 2.92
4.80/(3/4) = 6.4 per pound
8.2 * 6.4 = $52.48
Solution: $52.48 for 8.2 pounds
Answer:
1114$
Step-by-step explanation: