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BabaBlast [244]
3 years ago
10

Is AB congruent to DG

Mathematics
2 answers:
topjm [15]3 years ago
5 0
No the are not the same length
Ne4ueva [31]3 years ago
3 0
Yes they are congruent.
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Please help!!
Sergeu [11.5K]

The answer is A,you can refer to the definition

5 0
3 years ago
You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not a
11111nata11111 [884]

Answer:

For the exponential distribution:

\mu = \frac{1}{\lambda}

\sigma^2 = \frac{1}{\lambda^2}

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

\bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = \bar X

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

For this case we have a large sample size n =60 >30

The exponential distribution is the probability distribution that describes the time between events in a Poisson process.

For the exponential distribution:

\mu = \frac{1}{\lambda}

\sigma^2 = \frac{1}{\lambda^2}

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

\bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = \bar X

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}

5 0
4 years ago
Please answer full question below thank you
Yuliya22 [10]

Explanation:

Angles 1 & 2 have to equal 180. So, if ∠1 is equal to 140, ∠2 is equal to 40 because 140 + 40 = 180.

So far... we know that:

  • ∠2 = 40
  • ∠9 = 80

Angles 2, 9, and 11 make a triangle, a triangles "magic number" is also 180. To get 180, you must add up all of the angles. Well... we don't know ∠11 so to find it, you subtract 40 & 80 from 180. 180 - 80 - 40 = 60.

Answer:

∠2 = 40

∠11 = 60

7 0
4 years ago
What is the missing length?<br> с<br> 4 mm<br> 3 mm<br> 2<br> area = 10 mm
Anarel [89]

Answer:

4mm

Step-by-step explanation:

That is the answer thank you hehehe

3 0
3 years ago
Find the value of x.<br> x= 2.<br> x= 3<br> x= 33<br> x= 52
pshichka [43]

Answer: x=2

57+x=25x

subtract x from both sides

57+x-x=25x-x

57=24x

divide by 24 on both sides

57÷24=24x÷24

x=2.375

I hope this is good enough:

6 0
3 years ago
Read 2 more answers
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