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GenaCL600 [577]
3 years ago
10

We need to attach a support wire to Glendale's flagpole, which is 24 feet high. The

Mathematics
1 answer:
antoniya [11.8K]3 years ago
4 0

Step-by-step explanation:

this creates a right-angled triangle.

the right angle (90°) being the angle between flag pole and ground.

the angle ground-wire is 50°.

because we know that the sum of all angles in a triangle is always 180°, we know then that the angle flagpole-wire is

180 - 90 - 50 = 40°

so, we have the situation to know 1 side and all angles.

to get the other sides we use best the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

where the sides and related angles are always opposite of each other.

in our case now we have

24/sin(50) = wire/sin(90) = wire/1 = wire = 31.32977494... ft

so, the rounded answer is that the wire must be

31.3 ft long

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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

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

            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} } } ]

                                      = [ 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.

6 0
3 years ago
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