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marin [14]
3 years ago
6

Simone has 5 in her flower shop.Each employee works 6 4/15 hours a day how many hours in total do the 5 empoyees work per day?

Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0
To figure out the total amount of hours an employee works, we have to multiply the total amount of employees by the amount of hours each employee would work. 

To simplify the mixed fraction of hours they work, convert the mixed fraction into an improper fraction:

6 \frac{4}{15}

Multiply the whole number by the denominator:

6 \times 15 = 90

Add this number to the numerator to convert the mixed fraction into an improper fraction:

90 + 4 = 94

6 \frac{4}{15} = \frac{94}{15}

Now we can multiply the two numbers together (note that a whole number can be treated as the number above the denominator 1):

5 employees working 94/15 hours a day:

\frac{5}{1} \times \frac{94}{15} = \frac{470}{15}

To simplify this fraction, divide the numerator by the denominator:

470 \div 15 = 31 R 5

The whole number stays as a number, and the remainder is the new numerator:

31 \frac{5}{15}

The fraction can be further simplified by dividing both the numerator and denominator by the greatest common factor:

Factors of 5: {1,5}
Factors of 15: {1,3,5,15}

\frac{5}{15} \div \frac{5}{5} = \frac{1}{3}

31 \frac{5}{15} = 31 \frac{1}{3}

The 5 employees will work a total of 31 and 1/3 hours a day.
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Step-by-step explanation:

5 is one of the numbers in 1,2,3,4,5,6,7,8

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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
NEED HELP! DUE IN 1 HOUR PLEASE
Alenkasestr [34]

Answer:

Step-by-step explanation

if my answer is wrong i'm sorry here is a manual on how to do it if its wrong

i think they are right but just in case

Identify the output values. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.

1 relation

2 function

3 relation

4 function

i know 4 is right but i don't know about others double check i suggest just in case i think they are right

6 0
3 years ago
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