Answer:
A. Similar using SAS; ∆ABC ~ ∆DFE
Step-by-step explanation:
m<B of ∆ABC = m<F of ∆DFE
AB corresponds to DF
AB/DF = 12/8 = 3/2
BC corresponds to FE
BC/FE = 24/16 = 3/2
Thus, the ratio of the corresponding sides of both triangles are the same. Therefore, both triangles has two sides that are proportional to each other, and also had an included corresponding angles that are congruent to each other.
By the SAS criterion, we can conclude that both triangles are similar to each other. That is, ∆ABC ~ ∆DFE
Answer:
Weights of at least 340.1 are in the highest 20%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

a. Highest 20 percent
At least X
100-20 = 80
So X is the 80th percentile, which is X when Z has a pvalue of 0.8. So X when Z = 0.842.




Weights of at least 340.1 are in the highest 20%.
Answer:5 by 8 by 5 by 8
Step-by-step explanation: the perimeter is 26 and one side if 5 so you know that another side is also 5. So add 5 and 5 together you get 10. Then subtract 10 from the perimeter(witch we know is 26) and you end up with 16. we know that the last two side lengths are the same so divide 16 by two and the sidle length is 8.
Answer:
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Step-by-step explanation:
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Answer:
Your teacher is right, there is not enough info
Step-by-step explanation:
<h3>Question 1</h3>
We can see that RS is divided by half
The PQ is not indicated as perpendicular to RS or RQ is not indicates same as QS
So P is not on the perpendicular bisector of RS
<h3>Question 2</h3>
We can see that PD⊥DE and PF⊥FE
There is no indication that PD = PF or ∠DEP ≅ FEP
So PE is not the angle bisector of ∠DEF