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2 years ago

Which angle is an adjacent interior angle to ∠JKM? ∠JKL ∠MKL ∠KLM ∠LMK

Mathematics

2 answers:

rosijanka [135]2 years ago

9 0

<u>Answer:</u>

∠MKL

<u>Step-by-step explanation:</u>

Two angles are said to be adjacent angles if they share a common vertex and a common side but they do not overlap.

The triangle given here is ΔMKL with a line extended outside from K to J.

The two angles ∠MKL and ∠MKJ are adjacent angles as they share a common vertex K and a common side which is MK. The angle ∠MKL comes in the closed triangle while ∠MKJ is formed by extending the point K.

Therefore, ∠MKL is an interior adjacent angle while ∠MKJ is the exterior adjacent angle here.

ratelena [41]2 years ago

5 0

Two angles are adjacent when they have a common side and a common vertex

<JKM and <MKL have a common side MK and a common vertex K

Answer is ∠MKL

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Step-by-step explanation:

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And we can find the usual limits with:

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And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

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And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

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stealth61 [152]

Answer: First option is correct.

Step-by-step explanation:

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Since we know that

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since we can see that sum of residual is more than 0.

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