The rest of the question is the attached figure.
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Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17

The correct answer is option D. WB = 17

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2. An exam readiness awareness taskforce at a Fremont National university sampled 200 students after the midterm to ask them whe

k0ka [10]

Answer:

c) (40+60+25)/200 or 63%

Step-by-step explanation:

n= 200 students

Did Well on the Midterm and Studied for the Midterm = 75

Did Well on the Midterm and Went Partying = 40

Did Poorly on the Midterm and Studied for the Midterm = 25

Did Poorly on the Midterm and Went Partying = 60

The number of students that did poorly on the midterm or went partying the weekend before the midterm is given by the sum of all students who did poorly to all students who went partying minus the number of students who did Poorly on the Midterm and Went Partying:

$N=(25+60)+(40+60) - 60\\N= 125$