1.QUAD is an isosceles trapezoid 1. Given
2. QD = UA 2. Opposite legs of an isosceles trapezoid are equal
3. QU = QU 3. Reflexive
4. QA = UD 4. Diagonals of an isosceles trapezoid are equal.
5. ΔUAQ = ΔQDU 5. SSS
6. ∠UAQ = ∠QDU 6. CPCTE
Step-by-step explanation:
This is the answer.... Working shown
Answer:
Step-by-step explanation:
Complete Question
In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.
Answer:
16.5°
Step-by-step explanation:
In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.
We solve using Sine rule formula
a/sin A = b/sin B
We are solving for angle W
∠V=136°
Hence:
22 /sin 136 = 9 /sin W
Cross Multiply
22 × sin W = sin 136 × 9
sin W = sin 136 × 9/22
W = arc sin [sin 136 × 9/2.2]
W = 16.50975°
W = 16.5°
Answer:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:

And replacing we got:

Now we can find the second moment given by:

And replacing we got:

And the variance would be given by:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:
