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Usimov [2.4K]
3 years ago
15

Which of the following completes the proof? Triangles ABC and EDC are formed from segments BD and AC, in which point C is betwee

n points B and D and point E is between points A and C. Given: Segment AC is perpendicular to segment BD Prove: ΔACB ~ ΔECD Reflect ΔECD over segment AC. This establishes that ________. Then, ________. This establishes that ∠E'D'C' ≅ ∠ABC. Therefore, ΔACB ~ ΔECD by the AA similarity postulate. ∠ABC ≅ ∠E'D'C'; translate point E' to point A ∠ACB ≅ ∠E'C'D'; translate point E' to point B ∠ACB ≅ ∠E'C'D'; translate point D' to point B ∠ABC ≅ ∠E'D'C'; translate point D' to point A
Mathematics
2 answers:
skelet666 [1.2K]3 years ago
4 0

Answer:

∠ACB ≅ ∠E'C'D'. translate point E' to point A

Step-by-step explanation:

kotykmax [81]3 years ago
3 0

Answer: ∠ACB ≅ ∠E'C'D'; translate point D' to point B

Step-by-step explanation:

That is just my best guess.

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Help pleaseeeeeee, I need help
Arlecino [84]

Answer:

15, 60, and 61

Step-by-step explanation:

First we have to find the 3rd side length from 2 side length chosen. So let’s pick 8 meters and 11 meters, to find the third side length we use the Pythagorean theorem: c^2 = a^2 + b^2

c = 11^2 + 8^2 = 121 + 64 = 185 = 13.6

13.6 isn’t one of the answers so we can’t pick 11 & 8. Let’s try 11 & 15:

c = 15^2 + 11^2 = 225 + 121 = 346 = 18.6

18.6 isn’t one of the answers either. Let’s try 15 & 60:

c = 60^2 + 15^2 = 3600 + 225 = 3825 = 61.85

61.85 might be it but let’s try one more just in case:

c = 60^2 + 61^2 = 3600 + 3721 = 7321 = 85.6

Nope! That means 15, 60, and 61 are the answer. I hope this is correct!

5 0
3 years ago
Round each number to the nearest hundredth.
Nina [5.8K]
A. 8.75
B. 0.72
C. 9.31
D. 23.04
E. 6.98
5 0
3 years ago
Read 2 more answers
Bob and Tom each have a bag of 100 chocolate candies. Bob has 53 left and Tom has
arlik [135]

Answer:

He has 47

Step-by-step explanation:

if one guy has 50 it wold make sense to give the other 50 but since he has 53 then he must have 47

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3 years ago
ind the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximatio
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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
salantis [7]

Answer:

With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of​$131.00. Assume the population standard deviation is​$18.80.

  • Firstly, the pivotal quantity for 90% confidence interval for the  population mean is given by;

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

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                   of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.645 \times {\frac{18.8}{\sqrt{60} } } , 131+1.645 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

  • Now, the pivotal quantity for 95% confidence interval for the  population mean is given by;

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

where, \bar X = sample mean price = $131

            \sigma = population standard deviation = $18.80

            n = sample of home theater = 60

            \mu = population mean

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                                  = [131-1.96 \times {\frac{18.8}{\sqrt{60} } } , 131+1.96 \times {\frac{18.8}{\sqrt{60} } } ]

                                                  = [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The ​95% confidence interval is wider than the​ 90%.

7 0
3 years ago
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