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Alexxx [7]
3 years ago
6

My neighbor has an automatic sprinkler system for his front yard. It is activated for 10 minutes each of 2 sessions. There are 1

0 sprinkler heads each providing 1.5 gallons of water per minute. How many gallons of water is the neighbor using during a 30 day billing cycle?
Advanced Placement (AP)
1 answer:
marta [7]3 years ago
7 0

Answer: 9000

Given:

10 automatic sprinkler activated 10 minutes for 2 sessions daily.

Each sprinkler provides 1.5 gallons of water per minute

Question:

How many gallons of water is the neighbor using during a 30-day billing cycle?

Solution:

Each sprinkler provides water (10*2) or 20 minutes daily. In 20 minutes each sprinkler can provide (20*1.5) or 30 gallons of water. There are 10 sprinklers so (10*30) 300 gallons all in all daily. In 30 days, (30*300)9000 gallons of water are consumed.


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The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less is 0.0020

<h3>What is probability? </h3>

Probability can be defined as the likelihood of an event to occur. In statistics, the mean of the sample distribution typically shows the probability of the population.

From the parameters given:

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The sample proportion \mathbf{\hat p} can be computed by using the expression:

\mathbf{\hat p = \dfrac{x}{n}}

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\mathbf{\hat p =0.582}

If the percentage of the probability of all patients in the emergency room = 0.75

The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less can be computed as:

\mathbf{( \hat P \leq 0.582) = P \Big( \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}} \leq \dfrac{0.582 - 0.75}{\sqrt{\dfrac{0.75(1-0.75)}{55}}} \Big )}

\mathbf{( \hat P \leq 0.582 )= P \Big( Z \leq \dfrac{-0.168}{\sqrt{0.003409}} \Big )}

\mathbf{( \hat P \leq 0.582) = P \Big( Z \leq -2.88 \Big )}

From the Z distribution table:

\mathbf{( \hat P \leq 0.582) = 0.00198}

\mathbf{( \hat P \leq 0.582) \simeq 0.0020}

Learn more about probability here:

brainly.com/question/24756209

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