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vichka [17]
1 year ago
6

what do you do in the following problem... "Adam, a suburb dweller, decides to have a garden in the backyard. It is a rectangle

with a length of 24 feet and a width of 14 feet. He wants to divide the Garden in half along the diagonal. How many feet of fence does he need?"
Mathematics
1 answer:
andre [41]1 year ago
4 0

The opposite sides of a rectangle are equal. The diagram of the triangular garden is shown below:

The diagonal divides the garden into two equal parts. We would determine the length of the diagonal, d. We would apply pythagoras theorem which is expressed as

hypotenuse^2 = opposite side^2 + adjacent side^2

\begin{gathered} d^2=24^2+14^2 \\ d^2=576\text{ + 196 = 772} \\ d\text{ = }\sqrt[]{772\text{ }} \\ d\text{ = 27.79} \end{gathered}

27.79 feet of fence is needed

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The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt
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Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                             P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average height = 175 cm

            \sigma = population standard deviation = 20 cm

            n = sample of men = 40

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

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

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                     level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times }{\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for </u>\mu = [ \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 175-1.96 \times }{\frac{20}{\sqrt{40} } } , 175+1.96 \times }{\frac{20}{\sqrt{40} } } ]

                                            = [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

<em>The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.</em>

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