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victus00 [196]
3 years ago
9

Are the angle measurements in this triangle correct?

Mathematics
2 answers:
omeli [17]3 years ago
8 0

Answer:

can i see the triangle?

Step-by-step explanation:

nordsb [41]3 years ago
6 0
No the top angle should be 120°
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Each padlock is $3.90
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Need help on geomtry homework
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Perimeter = 42 cm

Step-by-step explanation:

XY = 8 cm  because ΔWXZ ≅ ΔYZX

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7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that
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Answer:

a)

<em>The probability that a single score drawn at random will be greater than 110  </em>

<em>P( X > 110) = 0.1587</em>

<em>b) </em>

<em>The probability that a sample of 25 scores will have a mean greater than 105</em>

<em>  P( x> 105) = 0.0062</em>

<em>c) </em>

<em>The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105)  = 0.002</em>

<em></em>

<em>d) </em>

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em>    P( 95 ≤ X≤ 105) = 0.9544</em>

<em></em>

Step-by-step explanation:

<u><em>a)</em></u>

Given mean of the Normal distribution 'μ'  = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1

<em>The probability that a single score drawn at random will be greater than 110</em>

<em>P( X > 110) = P( Z >1)</em>

                = 1 - P( Z < 1)

               =  1 - ( 0.5 +A(1))

               = 0.5 - A(1)

               = 0.5 -0.3413

              = 0.1587

b)

let 'X' = 105

Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5

<em>The probability that a single score drawn at random will be greater than 110</em>

<em>  P( x> 105) = P( z > 2.5)</em>

<em>                    = 1 - P( Z< 2.5)</em>

<em>                    = 1 - ( 0.5 + A( 2.5))</em>

<em>                   = 0.5 - A ( 2.5)</em>

<em>                  = 0.5 - 0.4938</em>

<em>                  = 0.0062</em>

<em>The probability that a single score drawn at random will be greater than 105</em>

<em>  P( x> 105) = 0.0062</em>

<em>c) </em>

let 'X' = 105

Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } =  4

<em>The probability that a single score drawn at random will have a mean greater than 105</em>

<em>  P( x> 105) = P( z > 4)</em>

<em>                    = 1 - P( Z< 4)</em>

<em>                    = 1 - ( 0.5 + A( 4))</em>

<em>                   = 0.5 - A ( 4)</em>

<em>                  = 0.5 - 0.498</em>

<em>                  = 0.002</em>

<em> The probability that a sample of 64 scores will have a mean greater than 105</em>

<em> P( x⁻> 105)  = 0.002</em>

<em>d) </em>

<em>Let  x₁ = 95</em>

Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } =  -2

<em>Let  x₂ = 105</em>

Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } =  2

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

                         = P(z≤2) - P(z≤-2)

                        = 0.5 + A( 2) - ( 0.5 - A( -2))

                      = A( 2) + A(-2)       (∵A(-2) =A(2)

                     =  A( 2) + A(2)  

                    = 2 × A(2)

                  = 2×0.4772

                  = 0.9544

<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>

<em>    P( 95 ≤ X≤ 105) = 0.9544</em>

<em>    </em>

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