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dolphi86 [110]
3 years ago
11

A ladder leans against a building that angle of elevation of the latter is 70° the top of the ladder is 25 feet from the ground.

to the nearest 10th of a foot how far from the building is the base of the ladder a. 20.5 feet b. 30.5 feet C.32.3’ or D.39.5 feet
Mathematics
2 answers:
Aloiza [94]3 years ago
5 0

<u>Answer:</u>

The correct answer option is a. 20.5 feet.

<u>Step-by-step explanation:</u>

We are given that the angle of elevation of the ladder is 70° and the height of the ladder is 25 feet from the ground.

We are to find the distance of the building from the base of the ladder.

For this, we will use tan:

tan 70 = \frac { 2 5 } { x }

x = \frac { 2 5 } { tan 7 0 }

x = 20.5 feet

VashaNatasha [74]3 years ago
3 0

Answer:

a. 20.5

Step-by-step explanation:

because this will form a right triangle we can use tan (opposite over adjacent) so an equation we could set up would be tan(70)=25/x

therefore we can just solve the equation which would give us 20.45. so if we round it the answer would be a

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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.

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Answer:

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Step-by-step explanation:

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