Write the next three terms of the arithmetic sequence.
2 answers:
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We have to prove that rectangles are parallelograms with congruent Diagonals.
Solution:
1. ∠R=∠E=∠C=∠T=90°
2. ER= CT, EC ║RT
3. Diagonals E T and C R are drawn.
4. Shows Quadrilateral R E CT is a Rectangle.→→[Because if in a Quadrilateral One pair of Opposite sides are equal and parallel and each of the interior angle is right angle than it is a Rectangle.]
5. Quadrilateral RECT is a Parallelogram.→→[If in a Quadrilateral one pair of opposite sides are equal and parallel then it is a Parallelogram]
6. In Δ ERT and Δ CTR
(a) ER= CT→→[Opposite sides of parallelogram]
(b) ∠R + ∠T= 90° + 90°=180°→→→Because RECT is a rectangle, so ∠R=∠T=90°]
(c) Side TR is Common.
So, Δ ERT ≅ Δ CTR→→[SAS]
Diagonal ET= Diagonal CR →→→[CPCTC]
In step 6, while proving Δ E RT ≅ Δ CTR, we have used
(b) ∠R + ∠T= 90° + 90°=180°→→→Because RECT is a rectangle, so ∠R=∠T=90°]
Here we have used ,Option (D) : Same-Side Interior Angles Theorem, which states that Sum of interior angles on same side of Transversal is supplementary.
Answer:
V(X)= 39.10
V(Y)= 40
Step-by-step explanation:
Given that
Total number of student = 140
Bus A - 31
Bus B- 43
Bus C- 27
Bus D- 39
The probability that a student was on the bus is proportional to the number of student. Eg 31/140 in bus A, 43/140 on bus B, ...
E(X) = (31*31/140) + (43*43/140) + (27*27/140) + (39*39/140)
= 35.5
V(X)= 39.10
The bus driver have 1/4 probability on being on any of the buses.
E(Y) = 140/4 = 35
V(Y)= 40