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dusya [7]
3 years ago
15

Find the area of the triangle below??

Mathematics
1 answer:
Sladkaya [172]3 years ago
8 0

Answer:

94.8

Step-by-step explanation:

94.8 is the answer for 1st q

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533/938 = .56 = 56%

Step-by-step explanation:

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Step-by-step explanation: You can represent that as 10^5*10^-5

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Q4.
goblinko [34]

The coding of the statistic is used to make it easier to work with the large sunshine data set

  • The mean of the sunshine is 3.05\overline 6
  • The standard deviation is approximately  <u>18.184</u>

<u />

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

Mean, \ \overline x = \dfrac{\sum x_i}{n}

The mean of the daily total sunshine is therefore;

Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6

s = \dfrac{x}{10 } - \dfrac{1}{10}

  • E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}

E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6

  • The mean ≈ 3.05\overline 6

Alternatively

,The \ mean \  of \  the \  daily  \ total  \ sunshine,  \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6

The mean of the daily total sunshine, \overline s ≈3.05\overline 6

  • Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)

Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)

Therefore;

Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537

Therefore;

  • s = \sqrt{330.6537} \approx 18.184

The standard deviation, s_s ≈ <u>18.184</u>

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