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lutik1710 [3]
3 years ago
11

rom a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Ve

gas. What is the probability that Marnie and Noomi are both selected?
Mathematics
1 answer:
andrezito [222]3 years ago
5 0

Answer:

probability that Marnie and Noomi are both selected=0.1

Step-by-step explanation:

A group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas.

the possibility of selecting 2 from 5 manages is 5C2 ways

5C2= \frac{5!}{2!(5-2)!} =10

There is only one choice of selecting Marine and Noomi from 5 managers

So probability of selecting Marine and Noomi is

\frac{1}{10} =0.1

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Write each equation in slope-intercept form.
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Answer:

A. y = -3/2x + 1

B. y = -5/3x + 10

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Step-by-step explanation:

A. You just have to isolate the y so divide by 4: y = -3/2x + 1

B. 3y = -5x + 30

Divide by 3: y = -5/3x + 10

C. -5x + 6 = 5y

Divide by 5: -1x + 6/5 = y OR y = -x + 6/5

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What is the property of yx7=7y
Artyom0805 [142]

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3 years ago
Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f
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Answer:

It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x) (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

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3 years ago
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