width=2
length= 4
explanation in the attached picture
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:
none, because without the letters and numbers.the plate number can not be identify or formed
As given he made 3/5 of basket he shot.
So we can find number of basket made by Everett equal to 3/5 of total number of shot .
So if total number of shot is 60.
Then the number of basket made by Everett =

of 60.

So the number of basket made by Everett in 60 shots = 36.
Answer:
D
Step-by-step explanation:
You have 4 chairs originally, added 2 and if you separate the tables you would get T=whatever the number is. i think