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mash [69]
3 years ago
12

Drag and drop the answers into the boxes to correctly complete the statement.

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
6 0

Did you ever find the answers?


kaheart [24]3 years ago
5 0

1. First transformation is rotation rotation of 180 about the origin. This transformation has a rule:

(x,y)→(-x,-y).

If points E(-2,-4), F(-1,-1), D(-2,-1) are vertices of triangle EFD, then

  • E(-2,-4)→E''(2,4),
  • F(-1,-1)→F''(1,1),
  • D(-2,-1)→D''(2,1).

2. Second transformation is translation 1 unit left with a rule

(x,y)→(x-1,y).

Then

  • E''(2,4)→E'(1,4),
  • F''(1,1)→F'(0,1),
  • D''(2,1)→D'(1,1).

Answer: 1st: rotation of 180 about the origin; 2nd: translation 1 unit left.

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Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce
34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

= 1 - P(z \leq -2)

Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

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How do I solve for x in the first two?
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Answer: X= 7.55cm

Step-by-step explanation:

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The domain of the relationship is [-2,6]

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