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suter [353]
3 years ago
12

The pH of randomly selected swimming pools follows a normal distribution with mean 7.4 and standard deviation 0.22. A pool shoul

d have a pH between 7.2 and 7.8 so that chlorine treatment is effective. What percentage of pools having an acceptable pH level
Mathematics
1 answer:
skad [1K]3 years ago
3 0

Answer:

78.383%

Step-by-step explanation:

The pH of randomly selected swimming pools follows a normal distribution with mean 7.4 and standard deviation 0.22. A pool should have a pH between 7.2 and 7.8 so that chlorine treatment is effective.

Using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

For x = 7.2

z = 7.2 - 7.4/0.22

z = -0.90909

Probability value from Z-Table:

P(x = 7.2) = 0.18165

For x = 7.8

z = 7.8 - 7.4/0.22

z = 1.81818

Probability alue from Z-Table:

P(x = 7.8) = 0.96548

The probability of pools having an acceptable pH level(7.2 and 7.8)

= P(x = 7.8) - P(x = 7.2)

= 0.96548 - 0.18165

= 0.78383

Converting to percentage

0.78383 × 100

= 78.383%

The percentage of pools having an acceptable pH level is 78.383%

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I think the correct equation is

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Domain is the set of x-values (i.e. units produced) that are feasible. This is all the positive integer values + 0, in case that you only consider that can produce whole units.

Range is the set of possible results for c(x), i.e. possible costs.

You can derive this from the fact that c(x) is a parabole and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape (it open upwards given that the cofficient of x^2 is positive). Also limit the costs to be positive.

You can substitute some values for x to help you, for example:

x      y

0    200

1    200 -7 +0.345 = 193.345

2    200 - 14 + .345 (4) = 187.38

3    200 - 21 + .345(9) = 182.105

4    200 - 28 + .345(16) = 177.52

5    200 - 35 + 0.345(25) = 173.625

6    200 - 42 + 0.345(36) = 170.42

10  200 - 70 + 0.345(100) =164.5

11 200 - 77 + 0.345(121) = 164.745

 

 

The functions does not have real roots, then the costs never decrease to 0.

The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.

Then the range goes to 164.5 to infinity, limited to the solutcion for x = positive integers.

4 0
3 years ago
0.285714285 nearest thousandth​
Agata [3.3K]

Answer:

0.286

Hope that helps!

Step-by-step explanation:

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B & C

Step-by-step explanation:

1.

sin C = \frac{AB}{AC} = \frac{12}{13}

2.

cosC=\frac{BC}{AC} = \frac{6}{10}

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David didn’t study for his introduction to logic exam consisting of 15 true-false questions. He did blind guessing on each quest
MakcuM [25]

Answer:

a) 0.1509

b) 0.8314

Step-by-step explanation:

Since there are only two options for each question, the probability that he guesses right is p = 0.50, and the probability that he guesses wrong is q = 0.50

There is a total of 15 questions in the test so we can use the formula for a binomial distribution with parameters n = 15, p = 0.5, q = 0.5

a)  If he needs to score 10 or more correct to pass, what is the probability that he will fail the exam?

To find this probability we will need to find the probability:

P ( X < 10) or

(1 -  P (x ≥10))

Remember that the formula for a binomial distribution is:

P(x) = \left[\begin{array}{ccc}n\\x\end{array}\right] p^{x} q^{n-x}

We can solve this by using the parameters given before and make x = 10, 11, 12, 13, 14, 15 and use the formula:

P(he fails the exam) = 1 - P (x≥10)

= 1 - (P(x=10) + P(x=11) + P(x=12) + P(x=13) + P(x=14) + P(x=15))

= 0.1509

b) Find the probability that he answers 6 to 11 inclusive)

We're going to use the same formula but we will do:

P (x = 6) + P (x = 7) + P (x = 8) + P (x = 9) + P (x = 10) + P (x = 11)

=0.1527 + .1964 + .1964 + 0.1527 + 0.0916 + 0.0416

= 0.8314

Therefore the probability that he answers 6 to 11 inclusive is 0.8314

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3 years ago
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