Answer:
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Proportional relationships always have the same outcome no matter how many batches are used. For example, scaling up or down a r
1 answer:
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Answer:
Step-by-step explanation:
a. The probability of selecting a 6 from the first draw and a 7 on the second draw when two balls are selected without replacement from a container with 10 balls numbered 1 to 10
Not independent because without replacement
Prob for both =
b. The probability of selecting a 6 on the first draw and a 7 on the second draw when two balls are selected with replacement from a container with 10 balls numbered 1 to 10
Here independent because with replacement makes probability independent.
Prob for both = P(A) *P(B) = d
c. The probability that two people selected at random in a shopping mall on a very busy Saturday both have a birthday in the month of June. Assume that all 365 birthdays are equally
likely, and ignore the possibility of a February 29 leap-year birthday.
Here independent because one person birthday will not affect the other person birthday
Prob for both =
d. The probability that two socks selected at random from a drawer containing 10 black socks and 6 white socks will both be black
Prob for I sock black = 10/16 and II sock black if first sock is black = 9/15
Hence not independent
Prob for both =
Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Answer:
y =
Step-by-step explanation:
with the point slope form, plug -9 (x coordinate) into x1, -3 (y coordinate) into y2, and -4/3 (slope) into m
y - y2 = m(x - x1)
y - (-3) = -4/3 (x - (-9))
y + 3 = -4/3 (x + 9)
y + 3 = -4/3 x + -12
y = -4/3 x - 12 - 3
y =