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Soloha48 [4]
1 year ago
15

a bag contains 3 red marbles 4 green marbles and 2 black marbles. what is the probability of a red marble being pulled from the

bag and without replacing it , the pulling a green marble out ?
Mathematics
1 answer:
zheka24 [161]1 year ago
3 0

Answer:

<u>1/6</u>

Step-by-step explanation:

<u>P (red)</u>

  • No. of red / Total
  • 3 / 3 + 4 + 2
  • 3/9
  • 1/3

<u>P (green without replacing red)</u>

  • No. of green / Total - 1
  • 4 / 9 - 1
  • 4/8
  • 1/2

<u>P (final)</u>

  • P (red) × P (green without replacing red)
  • 1/3 × 1/2
  • <u>1/6</u>
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Simplify this expression 3x +2a +5x +4a
umka21 [38]

Answer: 8x + 6a

Step-by-step explanation: Combine your "x" terms and your "a" terms.

So 3x + 5x is 8x and 2a + 4a is 6a.

So the answer is 8x + 6a.

5 0
3 years ago
Answer the following question about the function whose derivative is given below
Komok [63]

Answer:

a) The critical points are x = 3 and x = -6.

b) f is decreasing in the interval (-\infty, -6)

f is increasing in the intervals (-6,3) and (3,\infty).

c) Local minima: x = -6

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which f^{\prime}(x) = 0. So

f^{\prime}(x) = 0

(x-3)^{2}(x+6) = 0

So, the critical points are x = 3 and x = -6.

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if f^{\prime} is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are x = 3 and x = -6. So we have those following intervals:

(-\infty, -6), (-6,3), (3, \infty)

We select a point x in each interval, and calculate f^{\prime}(x).

So

-------------------------

(-\infty, -6)

f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100

f is decreasing in the interval (-\infty, -6)

---------------------------

(-6,3)

f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8

f is increasing in the interval (-6,3).

------------------------------

(3, \infty)

f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10

f is increasing in the interval (3,\infty).

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At x = -6, f goes from decreasing to increasing.

So x = -6, f assume a local minima value

At x = 3, f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

4 0
3 years ago
Elisa is writing an algebraic expression for the phrase 5 less than a number.Find her mistake and correct it
Arlecino [84]

Answer:

Let no be x

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Step-by-step explanation:


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lilavasa [31]

Answer:

16(0.5x – 0.75y + 2)

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3 0
2 years ago
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bixtya [17]
Answer=303.6

When rounding numbers to the nearest tenth, we need to look to the number in the hundredths place.

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If the number is 5 or greater, we round up. If the number is 4 or less, we round down. The number IS 5, so we round up.

303.55 becomes 303.6
3 0
3 years ago
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