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kkurt [141]
3 years ago
7

What is the probability of randomly selecting a z-score less than z = 1.25 from a normal distribution?

Mathematics
1 answer:
DedPeter [7]3 years ago
8 0

Answer:

The probability is 0.8944

Step-by-step explanation:

The probability of Z<1.25 in a normal distribution is

P(Z<1.25) = 0.5 + P(Z<1.25)

P(Z<1.25) = 0.5 + 0.3944

P(Z<1.25) = 0.8944

Hence, the probability of randomly selecting a z-score less than z = 1.25 from a normal distribution is 0.8944

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Find the unknown angle measure by solving for the given variable.
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Answer:

10 1/2

Step-by-step explanation:

8 0
2 years ago
A system of linear inequalities is shown below.
Helga [31]
A) Isolate y in both inequalities

1) x + y ≥ 4 => y ≥ 4 - x
2) y < 2x - 3

B) Draw the lines for the following equalities:

1) y = 4 - x
2) y = 2x - 3

C) Shade the regions of solutions

1) The region that is over the line y = 4 - x
2) The region that is below the line y = 2x - 3

The solution is the intersection of both regions; this is the sector between both lines that is to the right of the intersection point, including the portion of the very line y = 4 - x and excluding the portion of the very line y = 2x - 3


5 0
3 years ago
a bottle contains 4 fluid ounces of medicine. About how many milliliters of medicine are in the bottle?
yarga [219]
There is 118.2941184 milliliters in four fluid ounces
4 0
3 years ago
38,477 rounded to the nearest thousand
Ganezh [65]

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Step-by-step explanation:

4 0
2 years ago
Classify ABC by its sides. Then determine whether it is a right triangle.
m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

            AB = \sqrt{(9-3)^{2}+(6-3)^{2}  } = \sqrt{6^{2}+3^{2}  } =\sqrt{45}

Given vertices are  B(6,9) and C( 6,-3)

       B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

             =  \sqrt{(-3-9)^{2}+(6-6)^{2}  } =\sqrt{12^{2} } = 12

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 AC = \sqrt{(6-3)^{2}+(-3-3)^{2}  } = \sqrt{9+36} = \sqrt{45}

AC² = AB²+BC²

45  = 45+144

 45  ≠ 189

∴Given Δ ABC is not a right angle triangle

 

5 0
3 years ago
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