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Tom [10]
3 years ago
8

Why is 4 < x < 3 not a true solution?

Mathematics
1 answer:
OleMash [197]3 years ago
6 0

Answer:

Becase 4 is always more than 3 value

Step-by-step explanation:

4 < x < 3 can easily said as x is btw 4 and 3

If x is really more than 4, there's no way x will be less than 3

Let's persume as I got more 4 pcs of cake and less 2 pcs of cake. Why would I got less than 2 pcs if I got more than 4 pcs of cake?  Not make sense, right?

The correct will be  x is more than 3 and less than 4

or 3 < x < 4

Hope that help :)

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C is the correct answer
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Use the information from the polling agency to determine the sample size needed to construct a 95% confidence interval with a ma
Ber [7]

Answer:

the sample size n= 771

sample proportion is p = 1/2

Step-by-step explanation:

<u>Step1:</u>-

marginal error formula is \frac{2S.D}{\sqrt{n} }

but if 'n' is sample size then n = \frac{1}{(margin error)^{2} }

Given margin error is 36% = 0.036

n= \frac{1}{(0.036)^2}

on simplification , we get n= 777 polling agencies

<u>Step 2</u>:-

Margin of error = \frac{2\sqrt{p(1-p)} }{\sqrt{n} }

substitute values , we get

\frac{36}{100}  = \frac{2\sqrt{p(1-p)} }{\sqrt{771} }

after simplification, we get

\sqrt{p(1-p)} = \frac{1}{2}

squaring on both sides, we get

p(1-p) = \frac{1}{4}

multiply 'p' inside

p-p^{2} =\frac{1}{4}

cross multiplication '4' and simplify

4p^{2} -4p +1 =0 ........(1)

finding 'p' value

The equation (1) is of the form (a-b)^{2} = a^{2} -2ab+b^{2}

(2p)^{2} -2(2p)(1)+1^{2} = (2p-1)^2

(2p-1)^2 =0

p=\frac{1}{2}

<u>conclusion</u>:-

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the sample proportion is p = \frac{1}{2}

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