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OlgaM077 [116]
3 years ago
5

The first five multiples for the numbers 4 and 6 are shown below. Multiples of 4: 4, 8, 12, 16, 20, . . . Multiples of 6: 6, 12,

18, 24, 30, . . . What is the least common multiple of 4 and 6?
Mathematics
2 answers:
inn [45]3 years ago
5 0

Answer: The least common multiple of 4 and 6 is 12 because this is the smallest positive integer that is divisible by both 4 and 6.

Step-by-step explanation:

  • The least common multiple of two integers m and n is the least positive integer that is divisible by both m and n.

We are given that :

The first five multiples for the numbers 4 and 6 are shown below.

\text{ Multiples of 4: 4, 8, 12, 16, 20, . . . }\\\\\text{Multiples of 6: 6, 12, 18, 24, 30, . . . }

We can see that from the multiples of 4 and 6 , the least common multiple of 4 and 6 =12 such that 12 is divisible by 4 and 6 .

lubasha [3.4K]3 years ago
5 0

Step-by-step explanation:

12 boom

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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)

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Sample 2 (alcohol-type concentrates )

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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

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

[-4.78; 0.38]

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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?

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