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jeyben [28]
3 years ago
14

A lawn company advertises that they can spread 7,500 square feet of grass In 2 1/2 hours find the number of square feet of grass

that can be spread per hour
Add work please
Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0
Ok in one hour the company can spread 3,000 square feet or sqft.

First we need to see which numbers make sense out of 7,500. First we remove the 500 and have 7,000, then we separate 3,000 and for each hour and we are left with 1,000 from here and the 500 we took out before. Now we have a total of 1,500 for the half hour which makes sense because if you add 1,500 for half an hour an 1,500 for another half an hour you get a total of 3,000 which is what we got for the hours above and if you add 3,000 which is for one hour and add 3,000 for the other hour or multiple 3,000 *2 which is the same thing as adding it twice and then add 1,500 for the half hour we get 7,500 total which means 3,000 I'd how much the company can spread in one hour.

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What do u think which is more Simple Interest or Compound Interest?
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When it comes to investing, compound interest is better since it allows funds to grow at a faster rate than they would in an account with a simple interest rate.

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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
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avanturin [10]

Answer:

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likoan [24]

Step-by-step explanation:

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Secondly, rearrange the values on the right to look just like the general equation.

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Answer:

Answer: 216 cm2 (square centimetres cm^{2}, in your question you had to put cm3, cubic centimetres, it's IMPORTANT )

Step-by-step explanation:

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But let's calculate it anyway:

Volume  = Edge * Edge * Edge = length * width * depth = Edge^{3}  (remember: all edges are equal in this case)

so Edge = \sqrt[3]{Volume}

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Note: I call them "edges" but in case of a cube most say just "length"

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