Answer:
The answer is 90
Step-by-step explanation:
Using the dot product:
For any vector x, we have
||x|| = √(x • x)
This means that
||w|| = √(w • w)
… = √((u + z) • (u + z))
… = √((u • u) + (u • z) + (z • u) + (z • z))
… = √(||u||² + 2 (u • z) + ||z||²)
We have
u = ⟨2, 12⟩ ⇒ ||u|| = √(2² + 12²) = 2√37
z = ⟨-7, 5⟩ ⇒ ||z|| = √((-7)² + 5²) = √74
u • z = ⟨2, 12⟩ • ⟨-7, 5⟩ = -14 + 60 = 46
and so
||w|| = √((2√37)² + 2•46 + (√74)²)
… = √(4•37 + 2•46 + 74)
… = √314 ≈ 17.720
Alternatively, without mentioning the dot product,
w = u + z = ⟨2, 12⟩ + ⟨-7, 5⟩ = ⟨-5, 17⟩
and so
||w|| = √((-5)² + 17²) = √314 ≈ 17.720
Answer:
7x+21
Step-by-step explanation:
<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
