The part of the triangles which are congruent according to the description are; segment AB and segment DE.
<h3>Which parts of the triangles are congruent?</h3>
It follows from the task content that the two triangles ABC and DEF have been established as congruent. On this note, it can be established that by the congruence theorem that corresponding sides which are congruent and whose ratios equal to a constant ratio are segments AB and segment DE.
Read more on congruence theorem;
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Answer:
67.91
Step-by-step explanation:
Answer:
I think it's A or B
Step-by-step explanation:
I found the options online. I will state them and provide the answer.
OPTIONS
<span>A. find out what the prospective employer has to offer.
B. outline your value to a prospective employer.
C. accompany your portfolio.
D. remind the prospective employer of your recent interview.
</span>
ANSWER.
A letter of application is intended to outline your value to a prospective employer (option B) It is also known as a cover letter. It might be considered a job application which you send together with your CV to provide additional info about you.
Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 0°C and a standard deviation of 1.00°C.
This means that 
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69



has a p-value of 0.0455
X = -2.23



has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch: