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3 years ago

The value x = 3 is a solution to each of the following except which?

Mathematics

1 answer:

ololo11 [35]3 years ago

6 0

Answer:

$x \: is \: not \: a \: solution \: in \to \: 5x + 4 < 19. \\$

Step-by-step explanation:

$\: if \: x = 3 \: then \\ 4x-1=2x +5 \: is \: same \: as \to \: \\ 4(3) - 1 = 2(3) + 5\\ 12 - 1 = 6 + 5 \\ 11 =11 \to \: \\ \boxed{ x \: is \:a \: solution \: here} \\ \\ if \: x = 3 \\ then \to \\ 5x + 4 < 19\: is \: same \:as \to \\ 5(3) + 4 < 19 \\ 15 + 4 < 19 \\ 19 \: is \: equal \: to \: 19 : \\ 19 \: can \: not \: be \: less \: than \: 19 \\ \boxed{ henc e\: x \: is \: not \: a \: solution \: here}$

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3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

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Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

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