Triangle RTS is congruent to RQS by AAS postulate of congruent
Step-by-step explanation:
Let us revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles
and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
∵ SR bisects angle TSQ ⇒ given
∴ ∠TSR ≅ ∠QSR
∴ m∠TSR ≅ m∠QSR
∵ ∠T ≅ ∠Q ⇒ given
∴ m∠T ≅ m∠Q
In two triangles RTS and RQS
∵ m∠T ≅ m∠Q
∵ m∠TSR ≅ m∠QSR
∵ RS is a common side in the two triangle
- By using the 4th case above
∴ Δ RTS ≅ ΔRQS ⇒ AAS postulate
Triangle RTS is congruent to RQS by AAS postulate of congruent
Learn more:
You can learn more about the congruent in brainly.com/question/3202836
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Answer:
0.5 = 50% probability a value selected at random from this distribution is greater than 23
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:
What is the probability a value selected at random from this distribution is greater than 23?
This is 1 subtracted by the pvalue of Z when X = 23. So
has a pvalue of 0.5
0.5 = 50% probability a value selected at random from this distribution is greater than 23
Step-by-step explanation:
we can factor out z
so that'll be
z((1/m)-(1/n))=1
divide by the value in parentheses on both sides
z=1/((1/m)-(1/n))
I can't tell if there's more the problem wants you to do. I could rewrite the equation on paper more easily, but I don't have one on hand
Answer:
32
Step-by-step explanation:
The three consecutive numbers are: 32, 33, 34
They add up to 99
