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Dvinal [7]
2 years ago
15

N. -Y X 106° Please help I need this answer please.

Mathematics
2 answers:
brilliants [131]2 years ago
8 0

We need

  • m<NXY

Angles are supplementary

  • m<NXY+106=180
  • m<NXY=180-106
  • m<NXY=74
Harman [31]2 years ago
8 0

Answer:

74°

Step-by-step explanation:

A bearing is an angle, measured clockwise from the north direction.

To find the bearing of Y from X:

  • Start at point X
  • Measure clockwise from north to the line going to point Y.

Angles on a straight line sum to 180°

⇒ bearing of Y from X = 180° - 106°

                                     = 74°

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3 years ago
The mean per capita income is 23,037 dollars per annum with a variance of 149,769. What is the probability that the sample mean
mihalych1998 [28]

Answer:

Probability that the sample mean would be less than 23,013 dollars is 0.2358.

Step-by-step explanation:

We are given that the mean per capita income is 23,037 dollars per annum with a variance of 149,769.

Also, a sample of 134 persons is randomly selected.

Firstly, Let \bar X = sample mean

The z-score probability distribution for sample mean is given by;

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

where, \mu = population mean per capita income = 23,037 dollars p.a.

           \sigma = standard deviation = \sqrt{Variance} = \sqrt{149,769} = 387 dollars p.a

           n = sample of persons = 134

So, probability that the sample mean would be less than 23,013 dollars is given by = P(\bar X < 23,013 dollars)

      P(\bar X < 23,013) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{23,013-23,037}{\frac{387}{\sqrt{134} } } ) = P(Z < -0.72) = 1 - P(Z \leq 0.72)

                                                                    = 1 - 0.7642 = 0.2358

The above probability is calculated using the z-score table.

<em>Therefore, probability that the sample mean would be less than 23013 dollars if a sample of 134 persons is randomly selected is 0.2358.</em>

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