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GalinKa [24]
3 years ago
15

Which relationship is always correct for the angles p, q, and r of triangle ABC?

Mathematics
1 answer:
Verizon [17]3 years ago
7 0

Answer:

the right answer is

p + r = q

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Can someone pls check if my answer is accurate
Alekssandra [29.7K]

Answer:

yes

Step-by-step explanation:

3 0
4 years ago
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Find the value of c that makes z2 - 5z + c a perfect square trinomial.
pogonyaev

Answer:

  • c = 6.25

Step-by-step explanation:

<u>Given trinomial:</u>

  • z² - 5z + c

<u>Complete square:</u>

  • z² - 2*2.5*z + 2.5² + c - 2.5² =
  • (x - 2.5)² + c - 6.25

<u>The value of c:</u>

  • c - 6.25 = 0
  • c = 6.25
5 0
2 years ago
What is the gradient of the graph shown? <br><br> Give your answer in its simplest form.
gregori [183]

Answer:

nffjebfjeqbhq2bb cinderall kwebfjbw

Step-by-step explanation:

bsdbfhbshfbhfbwhfvcnncsh  jdhsshbfhwapv

3 0
3 years ago
How many pairs of parallel line segments are shown? <br><br> 1. 8<br> 2. 24<br> 3. 4<br> 4. 18
Burka [1]
For this case what you should see is each of the edges of the prism that are parallel.
 We then have as parallel edges:
 AC and GE
 CG and AE
 CD and GH
 AB and EF
 BD and HF
 DH and BF
 CD and EF
 GH and AB
 Answer: 
 8 pairs of parallel lines
8 0
4 years ago
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Which is the expected value of the random variable with given probability distribution ?
Gnom [1K]

Answer:

Step-by-step explanation:

Let X be the RV we are talking about.

Expected value has different aliases, such as weighted average, mean, expectation, first moment and so on.

Okay, how do we calculate the average first, just sum all the values and divide by number of values. It comes from the basic assumption, which is choosing one of them is equally likely. What I mean by that is quite simple, above process what I talked about is just sum of the numbers times their probabilities. e.g. (a+b+c+d)/4 = a/4 + b/4 + c/4 + d/4. In the example, we assumed that probability of choosing any a, b, c and d is equally likely.

Then, why I use the term weighted average is the generalized situation of average, which is some of the numbers are repetitive. For instance, X can take the values a, b, c and d; however, X can take the value a 3 times more than d, X can take the value b 2 times more than d and c and d occurs with the same frequency at X, Therefore, if we take a sample, it is likely to be a, b, a, c, d, a, b or the permutation of this sample. Therefore the weighted average is 3 * a / 7 + 2 * b / 7 + c / 7 + d /7 or using normal average theorem (a + a + a + b + b + c +d) / 7. First, both of them are equal. Find the distribution and multiply the values with the probabilities of the distribution and sum them up or just add them and divide by number of sample. Second, if you check that, a has more weight, which means a can determine the direction of the average more. That is the logic behind the expected value.

Now, if X is a discrete RV (like in the above example), we know what we are going to do. What if X is a continuous RV ? X could take infinite values therefore if you divide a number with an infinite value, the probability of an individual value X could take is going to be 0. Therefore, we cannot calculate the expected value of chosen values individually; however, we can calculate the expected value of an interval by approximating with Riemann Sum or with a better choice, using integrals. Because, technically using Riemann Sum and quantizing the values and approximating the area of the curve is the same as we did above.

In conclusion, basically if X is a discrete RV and the values X could take is countably infinite,

If X is a discrete RV and the values X could take is finite,

if X is a continuous RV,

3 0
3 years ago
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