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Galina-37 [17]
3 years ago
6

Ten percent of U.S. households contain 5 or more people. You want to simulate choosing a household at random and recording "Yes"

if it contains 5 or more people. Which of these are correct assignments of digits for this simulation?
(a) Odd = Yes; Even = No
(b) 0 = Yes; 1-9 = No
(c) 0-5 = Yes; 6-9 = No
(d) 0-4 = Yes; 5-9 = No
(e) None of these
Mathematics
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

  (b)  0 = Yes; 1-9 = No

Step-by-step explanation:

Among digits, 10% of 10 digits is a single digit. Any answer choice that suggests more than one digit be used for "yes" is not appropriate. A good choice is ...

  0 = Yes; 1-9 = No . . . . . choice (b)

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HK consists of two part, HJ and JK

HK = HJ + JK

We know that:

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Subtract 3.1ft from both sides.

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Select three ratio that are equivalent to 8:20
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Answer:

16:40 24:60 32:80

Step-by-step explanation:

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D.sqrt(2+x^/2)<br> Solve this question please
lidiya [134]

Answer:

Option a.

Step-by-step explanation:

By looking at the options, we can assume that the function y(x) is something like:

y = \sqrt{4 + a*x^2}

y' = (1/2)*\frac{1}{\sqrt{4 + a*x^2} }*(2*a*x) = \frac{a*x}{\sqrt{4 + a*x^2} }

such that, y(0) = √4 = 2, as expected.

Now, we want to have:

y' = \frac{x*y}{2 + x^2}

replacing y' and y we get:

\frac{a*x}{\sqrt{4 + a*x^2} } = \frac{x*\sqrt{4 + a*x^2} }{2 + x^2}

Now we can try to solve this for "a".

\frac{a*x}{\sqrt{4 + a*x^2} } = \frac{x*\sqrt{4 + a*x^2} }{2 + x^2}

If we multiply both sides by y(x), we get:

\frac{a*x}{\sqrt{4 + a*x^2} }*\sqrt{4 + a*x^2} = \frac{x*\sqrt{4 + a*x^2} }{2 + x^2}*\sqrt{4 + a*x^2}

a*x = \frac{x*(4 + a*x^2)}{2 + x^2}

We can remove the x factor in both numerators if we divide both sides by x, so we get:

a = \frac{4 + a*x^2}{2 + x^2}

Now we just need to isolate "a"

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2*a + a*x^2 = 4 + a*x^2

Now we can subtract a*x^2 in both sides to get:

2*a = 4\\a = 4/2 = 2

Then the solution is:

y = \sqrt{4 + 2*x^2}

The correct option is option a.

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Answer:

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