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balandron [24]
3 years ago
13

Help 30 points and i will mark brainliest for the best answer which will be more points

Mathematics
1 answer:
klasskru [66]3 years ago
6 0

Answer:

See below

Step-by-step explanation:

From the table we see there is a relation between x and y values

Every step change of x = 0.5, every step change of y = 2

<u>This relation is linear. The formula for linear function is:</u>

  • y = mx + b

<u>Using to pairs of points to get the slope and y-intercept</u>

  • (0, 60) and (1, 64)

<u>Calculating the slope:</u>

  • m = (64 - 60)/(1 - 0) = 4/1 = 4
  • y - intercept is 60 as per point (0, 60)

<u>So the function is</u>

  • y = 4x + 60
<h3>Part A</h3>
  • Yes. Described above
<h3>Part B</h3>
  • y = 4x + 60
<h3>Part C</h3>
  • Slope is the steepness of the line.
  • y -intercept is the lowest score with no time spent studying
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an Carlos County Health Department reported 17 new cases of tuberculosis in the first half of 2001 and 18 additional cases durin
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San Carlos County

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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

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            n = sample size = 16

            \mu = population mean

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

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

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

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

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