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katovenus [111]
3 years ago
10

The sum of two numbers is 60 and the difference is 2 what are the numbers.

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
Let the two numbers be x and y
Then
x + y = 60
And
x - y = 2
x = y + 2
Putting the value of x in the first equation we get
x + y = 60
y + 2 + y= 60
2y + 2 = 60
2y = 60 - 2
2y = 58
y = 58/2
y = 29
Now putting the value of y in the first equation we get
x + y = 60
x + 29 = 60
x = 60 - 29
x = 31
So the value of the two numbers comes out to be 31 and 29. I hope the procedure is clear enough for you to understand.
Kisachek [45]3 years ago
5 0
31 and 29 because 31  plus 29 is 60 and 31 minus 29 is 2
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enot [183]

Answer:

a)

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

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Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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The margin of error is:

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This means that \pi = 0.16

a. What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03 (round up to the next whole number).

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0.03\sqrt{n} = z\sqrt{0.16*0.84}

\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}

(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

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Question b:

99% confidence level,

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2

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A sample size of 991 is needed.

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\text{Hello! :D}

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