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Kipish [7]
3 years ago
14

Plz help me with dis its one question

Mathematics
1 answer:
tatiyna3 years ago
7 0

Answer:

Alix does. See below.

Step-by-step explanation:

Givens

<em><u>Jenny</u></em>

  • b1 = 90 cm
  • b2 = 80 cm
  • height = 23 cm

<em><u>Alex</u></em>

  • b1 = 90 cm
  • b2 = 70 cm
  • height = 27 cm

Formula

For both territories the answer is based on the formula

  • Area = (b1 + b2)*h/2

Solution

<em><u>Jenny</u></em>

  • Area = (90 + 80)23/2
  • Area = 170 * 23 / 2
  • Area = 85 * 23
  • Area = 1955 units.

<em><u>Alex</u></em>

  • Area = (70 + 90)*27/2
  • Area = 160*27 / 2
  • Area = 80* 27
  • Area = 2160

Alix does by 205 cm

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Compute the following: 325 C1
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A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point est
xxTIMURxx [149]

Answer:

Point estimate for the population variance = 3.92 * 10^{-3} .

Step-by-step explanation:

We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters ;

       X                       X - Xbar                                         (X-Xbar)^{2}

      1.13            1.13 - 1.188 = -0.058                                 3.364 * 10^{-3}

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      1.15            1.15 - 1.188 = -0.038                                 1.444 * 10^{-3}

      1.24           1.24 - 1.188 = 0.052                                 2.704 * 10^{-3}

      1.27           1.27 - 1.188 = 0.082                               <u>  6.724 * </u>10^{-3}<u>    </u>

                                                                    \sum (X-Xbar)^{2} <u>= 0.01568   </u>

Firstly, Mean of above data, Xbar = \frac{\sum X}{n} = \frac{1.15+1.24+1.15+1.27+1.13}{5} = 1.188

Point estimate of Population Variance = Sample variance

                                                               = \frac{\sum (X-Xbar)^{2}}{n-1} = \frac{0.01568}{4} = 3.92 * 10^{-3} .

Therefore, point estimate for the population variance = 3.92 * 10^{-3} .

       

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3 years ago
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