About what distance in feet would a person be 8 seconds after experiment begins

An article in Fire Technology investigated two different foam-expanding agents that can be used in the nozzles of firefighting s

UNO [17]

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= $\frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }$

a) 95% CI

(X[bar]_1 - X[bar]_2) ± $t_{n_1 + n_2 - 2}$ * $Sa* \sqrt{\frac{1}{n_1}+\frac{1}{n_2} }$

Sa²= $\frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 - 2}$ = $\frac{(5-1)0.36 + (5-1)0.64}{5 + 5 - 2}$ = 0.5

Sa= 0.707ç

$t_{n_1 + n_2 -2: 1 - \alpha /2} = t_{8; 0.975} = 2.306$

(4.7-6.9) ± 2.306* $(0.707\sqrt{\frac{1}{5}+\frac{1}{5} })$

[-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?

8 0

3 years ago

Expand the polynomial and simplify: (4xy− 1 8 y^2)^2

Vinvika [58]

Answer: $4y^{2}(2x-9y)^{2}$

Step-by-step explanation:

We have the following polynomial:

$(4xy-18y^{2})^{2}$

This is a polynomial of the form $(a-b)^{2}=a^{2}-2ab+b^{2}$ . Following this rule to expand it, we have:

$(4xy)^{2}-2(4xy)(18y^{2})+(18y^{2})^{2}$

$16x^{2}y^{2}-144xy^{3}+324y^{4}$

Applying common factor $4y^{2}$ :

$4y^{2}(4x^{2}-36xy+81y^{2})$

Note the polynomial inside the parenthesis is a perfect square trinomial, which can be factored to $(2x-9y)^{2}$ . Hence, the final simplification is:

$4y^{2}(2x-9y)^{2}$

3 0

2 years ago

A pen stand costs ₹120. How many pen stand can be bought for ₹1440

siniylev [52]

1440 divided by 120 = 12 pen stands

hope it helps:))

6 0

2 years ago

the first group of students consists of 10 and their average age was 13 years old the next group consisted of 30 students and th

Kazeer [188]

Answer: The average was 9 years old find the average age of both groups is 10 years old.

Step-by-step explanation:

Formula foe average : $\text{Average} =\dfrac{\text{Sum of all observations}}{\text{Number of observations}}$

Given : The first group of students consists of 10 and their average age was 13 years old.

i.e. $13=\dfrac{\text{Sum of ages of students of first group}}{10}\\\\\Rightarrow\ \text{Sum of ages of students of first group}=13\times10=130$ (1)

The next group consisted of 30 students and their average was 9 years old.

i.e. $9=\dfrac{\text{Sum of ages of students of next group}}{30}\\\\\Rightarrow\ \text{Sum of ages of students of nextgroup}=9\times30=270$ (2)

Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400

Combined students of both groups (first and next group )= 10+30=40

Now , the average of both groups = $\dfrac{\text{Sum of combined students}}{\text{combined students}}$

$=\dfrac{400}{40}=10$

Hence, the average was 9 years old find the average age of both groups is 10 years old.

4 0

3 years ago

Ayúdeme es para mañana

artcher [175]

Ok this is impossible you speak spanish

6 0

2 years ago