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nikdorinn [45]
3 years ago
12

Hydrologists collected water samples from a river for several years. They found 62 out of 648 samples exceeded the desired pH le

vels, with a 2% samples exceeded the desired pH levels, with a 90% confidence level. Which of the following is the most reasonable conclusion?
A. It is likely that the river water exceeds the desired pH level between 8% and 12% of the time studied.

B. It is likely that the river water exceeds the desired pH level between 10% and 12% of the time studied.

C. It is likely that the river water exceeds the desired pH level between 13% and 15% of the time studied.

D. It is likely that the river water exceeds the desired pH level between 11% and 15% of the time studied.
Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

Correct option is (A).

Step-by-step explanation:

Let <em>p</em> = proportion of water samples that exceeded the desired pH level.

A sample of size <em>n</em> = 648 is selected. Of these samples <em>X</em> = 62 exceeded the desired pH levels.

The confidence interval for the population proportion is given by:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\Rightarrow CI=\hat p\pm MOE

The MOE or margin of  error is estimated difference between the true population parameter value and the sample statistic value.

The information provided is:

MOE = 0.02

\hat p=\frac{X}{n}=\frac{6}{648}=0.096

Compute the 90% confidence interval for the proportion of water samples that exceeds the desired pH level as follows:

CI=\hat p\pm MOE\\=0.096\pm 0.02\\=(0.076, 0.116)\\\approx(8\%, 12\%)

Thus, the 90% confidence interval for the proportion of water samples that exceeds the desired pH level is (8%, 12%).

This confidence interval implies that there is a 90% confidence that the river water exceeds the desired pH level between 8% and 12% of the time studied.

The correct option is (A).

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Answer [83, 100]

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(3,3)

Step-by-step explanation:

Given

Vertices: (0,0)\ (1,0)\ (0,1)

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First, we need to determine the new vertices:

New\ Vertex = Scale\ Factor * Old\ Vertex

For (0,0):

New\ Vertex = 3 * (0,0)

New\ Vertex = (3 * 0,3 * 0)

New\ Vertex = (0,0)

For (1,0):

New\ Vertex = 3 * (1,0)

New\ Vertex =  (3 * 1,3 * 0)

New\ Vertex =  (3,0)

For (0,1):

New\ Vertex =  3 * (0,1)

New\ Vertex =  (3 * 0,3 * 1)

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5 0
3 years ago
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Answer:

P (Cheese) = 0.199, P (Sausage) = 0.259, P (Pepperoni) = 0.181,

P (Supreme) = 0.130, P (Another Kind) = 0.144

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Step-by-step explanation:

Given:

Number of students who prefer cheese = 43

Number of students who prefer sausage = 56

Number of students who prefer pepperoni = 39

Number of students who prefer supreme = 28

Number of students who prefer another kind = 31

Number of students who did not like any kind = 19

∴ The total number of students surveyed = 43+56+39+28+31+19=216       The number of students who prefer pizza = 43+56+39+28+31=197

The probability that a students likes pizza is,

P(Student\ likes\ pizza)=\frac{No.\ of\ students\ who\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}

                                     =\frac{197}{216} \\=0.912

The probability that a students does not likes pizza is,

P(Student\ does\ not\ likes\ pizza)=\frac{No.\ of\ students\ who\ does\ not\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}

                                                   =\frac{19}{216} \\=0.088

The probability distribution of students who prefer different kinds of pizza is:

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       P(A\ Student\ prefers\ cheese)=\frac{No.\ of\ students\ who\ prefer\ cheese}{Total\ no.\ of\ students\ surveyed}

                                                       =\frac{43}{216}\\=0.199

  • The probability that a student likes sausage:

        P(A\ Student\ prefers\ sausage)=\frac{No.\ of\ students\ who\ prefer\ sausage}{Total\ no.\ of\ students\ surveyed}

                                                           =\frac{56}{216}\\=0.259

  • The probability that a student likes pepperoni:

       P(A\ Student\ prefers\ pepperoni)=\frac{No.\ of\ students\ who\ prefer\ pepperoni}{Total\ no.\ of\ students\ surveyed}  

                                                             =\frac{39}{216}\\=0.181

  • The probability that a student likes supreme:

       P(A\ Student\ prefers\ supreme)=\frac{No.\ of\ students\ who\ prefer\ supreme}{Total\ no.\ of\ students\ surveyed}

                                                           =\frac{28}{216}\\=0.130

  • The probability that a student likes another kind:

        P(A\ Student\ prefers\ another\ kind)=\frac{No.\ of\ students\ who\ prefer\ another\ kind}{Total\ no.\ of\ students\ surveyed}

                                                                   =\frac{31}{216}\\=0.144

Thus, the probability distribution table is displayed below:

6 0
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