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Aliun [14]
2 years ago
14

A manufacturer claims that its cleaning product is best because over half of the cleaning services surveyed used the product. Wh

ich BEST decreases the validity of the claim? A.Only 50 cleaning services were surveyed. B.Cleaning services were selected at random. C.A well-known company conducted the survey. D.Only one cleaning product was asked about in the survey.
Mathematics
1 answer:
balu736 [363]2 years ago
8 0

Answer:

D. Only one cleaning product was asked about in the survey.

Any given company can use more than one cleaning product or even use a combination of different cleaning products. The question was not specific enough, since it should have probably asked which cleaning products are used and which ones are used more frequently.

Step-by-step explanation:

the other options are wrong because:

  • A. Only 50 cleaning services were surveyed. ⇒ This actually would increase the validity of the claim because it satisfies the the condition for a valid confidence interval [np and n(1 - p) are larger than 10]-
  • B. Cleaning services were selected at random. ⇒ This actually would increase the validity of the claim.
  • C. A well-known company conducted the survey. ⇒ This actually would increase the validity of the claim.

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Answer:

x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

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\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}

Second equation:

\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}

Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}

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Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

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