Answer:
(1) D.Angle C is congruent to to Angle F. (2) C. SSS. (3) C. cannot be congruent to.
Step-by-step explanation:
1)
From the given figure it is noticed that
According to SAS postulate, if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then both triangles are congruent.
The included angles of congruent sides are angle C and angle G.
So, condition "Angle C is congruent to to Angle F" will prove that the ∆ABC and ∆EFG are congruent by the SAS criterion.
2)
If 
According to SSS postulate, if all three sides in one triangle are congruent to the corresponding sides in the other.
Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore SSS criterion for congruence is violated.
3)
Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore the included angle of congruent sides are different.
Therefore angle C and angle F cannot be congruent to each other.
Hello.
We have:
- A point that the line passes through
- The slope of the line
Plug the values into the Point-Slope Formula:
I hope it helps.
Have a nice day.
3a plus 7. you group the whole numbers together and than the a's. so 3+4 is 7. 2a + a is 3a.
1. Total
2. Deficit
( hope this helps)
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
P(brown) = 12% = 0.12
P(Yellow) = 15% = 0.15
P(Red) = 12% = 0.12
P(blue) = 23% = 0.23
P(orange) = 23% = 0.23
P(green) = 15% = 0.15
A.) Compute the probability that a randomly selected peanut M&M is not yellow.
P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85
B.) Compute the probability that a randomly selected peanut M&M is brown or red.
P(Brown) or P(Red) :
0.12 + 0.12 = 0.24
C.) Compute the probability that three randomly selected peanut M&M’s are all brown.
P(brown) * P(brown) * P(brown)
0.12 * 0.12 * 0.12 =0.001728
D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.
P(3 blue)' = 1 - P(3 blue)
P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167
1 - P(3 blue) = 1 - 0.012167 = 0.987833
If you randomly select three peanut M&M’s, compute that probability that at least one of them is blue.
P(1 blue) + p(2 blue) + p(3 blue)
(0.23) + (0.23*0.23) + (0.23*0.23*0.23)
0.23 + 0.0529 + 0.012167
= 0.295067
