The given congruency of the sides 
and 
and 
and 
as well
as the congruency of the common side 
gives.
ΔBEL ≅ ΔLOB by SSS congruency postulate
<h3>Which values correctly completes the table?</h3>
The completed two column proof is presented as follows;
Statement 
Reasons
1. 
1. Given
2. 
≅ 
2.
3. 
3. <u>Reflexive property of congruency</u>
4. ΔBEL ≅ ΔLOB 4. <u>SSS congruency postulate</u>
Side-Side-Side, SSS, congruency postulate states that if three sides of
one triangle are congruent to three sides of another triangle, the two
triangles are congruent.
Learn more about different congruency postulates here:
brainly.com/question/1495556
Your answer to this question is B
To answer this question, you can use a factor tree, or the table thing (I forgot what it’s called.)
You divide the number by one of its PRIME factors, until there is only one left. The numbers you divided it by are written as shown.
Hope this helps. :)
Answer:
The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 3.25
The margin of error is:
M = T*s = 3.25*25 = 81
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 12,556 - 81 = 12,475 pounds
The upper end of the interval is the sample mean added to M. So it is 12,556 + 81 = 12,637 pounds.
The 99% confidence interval of the population mean for the weights of adult elephants is between 12,475 pounds and 12,637 pounds.
