To answer this
problem, we use the binomial distribution formula for probability:
P (x) = [n!
/ (n-x)! x!] p^x q^(n-x)
Where,
n = the
total number of test questions = 10
<span>x = the
total number of test questions to pass = >6</span>
p =
probability of success = 0.5
q =
probability of failure = 0.5
Given the
formula, let us calculate for the probabilities that the student will get at
least 6 correct questions by guessing.
P (6) = [10!
/ (4)! 6!] (0.5)^6 0.5^(4) = 0.205078
P (7) = [10!
/ (3)! 7!] (0.5)^7 0.5^(3) = 0.117188
P (8) = [10!
/ (2)! 8!] (0.5)^8 0.5^(2) = 0.043945
P (9) = [10!
/ (1)! 9!] (0.5)^9 0.5^(1) = 0.009766
P (10) = [10!
/ (0)! 10!] (0.5)^10 0.5^(0) = 0.000977
Total
Probability = 0.376953 = 0.38 = 38%
<span>There is a
38% chance the student will pass.</span>
company a
Step-by-step explanation:
if the work for company b takes more than two days is gonna be more than 1,000 dollars and if he pick company a he would owe 1,000 no matter how many days it takes
Answer:
20°
Step-by-step explanation:
40°, 70° and 90° are the measures of the three angles of the quadrilateral.
Measure of fourth angle of the Quadrilateral
= 360° - (40° + 70° + 90°)
= 360° - 200°
= 160°
Measure of angle 1 will be equal to the measure of the linear pair angle of 160° as they are corresponding angles.
Thus,


Alternate method:
![m\angle 1 = 180\degree- [360\degree-(40\degree+70\degree+90\degree)]](https://tex.z-dn.net/?f=m%5Cangle%201%20%3D%20180%5Cdegree-%20%5B360%5Cdegree-%2840%5Cdegree%2B70%5Cdegree%2B90%5Cdegree%29%5D)
![\implies m\angle 1 = 180\degree- [360\degree-200\degree]](https://tex.z-dn.net/?f=%5Cimplies%20m%5Cangle%201%20%3D%20180%5Cdegree-%20%5B360%5Cdegree-200%5Cdegree%5D)


The rest of the question is the attached figure.
============================================
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