Answer:
So, to find the residual I would subtract the predicted value from the measured value so for x-value 1 the residual would be 2 - 2.6 = -0.6.
Step-by-step explanation:
Answer:
<u>Identities used:</u>
- <em>1/cosθ = secθ</em>
- <em>1/sinθ = cosecθ</em>
- <em>sinθ/cosθ = tanθ</em>
- <em>cosθ/sinθ = cotθ</em>
- <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
- (1 - sinθ)/(1 + sinθ) =
- (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
- (1 - sinθ)² / (1 - sin²θ) =
- (1 - sinθ)² / cos²θ
<u>Square root of it is:</u>
- (1 - sinθ)/ cosθ =
- 1/cosθ - sinθ / cosθ =
- secθ - tanθ
<h3>Question 2 </h3>
<u>The first part without root:</u>
- (1 + cosθ) / (1 - cosθ) =
- (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
- (1 + cosθ)² / (1 - cos²θ) =
- (1 + cosθ)² / sin²θ
<u>Its square root is:</u>
- (1 + cosθ) / sinθ =
- 1/sinθ + cosθ/sinθ =
- cosecθ + cotθ
<u>The second part without root:</u>
- (1 - cosθ) / (1 + cosθ) =
- (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
- (1 - cosθ)²/ (1 - cos²θ) =
- (1 - cosθ)²/sin²θ
<u>Its square root is:</u>
- (1 - cosθ) / sinθ =
- 1/sinθ - cosθ / sinθ =
- cosecθ - cotθ
<u>Sum of the results:</u>
- cosecθ + cotθ + cosecθ - cotθ =
- 2cosecθ
Answer:
0.25
Step-by-step explanation:
This is the question on conditional probability
Let A - the quadrilateral has four right angles
B - the quadrilateral has four equal side lengths
Required probability = P(B/A)
By definition of continuous probability


P(A) = 
Hence given probability = 
<em><u>Hey</u></em><em><u>!</u></em><em><u>!</u></em>
<em><u>They</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>adjacent</u></em><em><u> </u></em><em><u>angles</u></em><em><u>.</u></em>
<em><u>Both</u></em><em><u> </u></em><em><u>angles</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>common</u></em><em><u> </u></em><em><u>side</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>vertex</u></em><em><u>.</u></em>
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>