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joja [24]
2 years ago
8

A randomly generated list of numbers from 0 to 4 is being used to simulate an event, with the number 4 representing a success. W

hat is the estimated probability of a success?
Mathematics
2 answers:
Sliva [168]2 years ago
7 0

Answer: 20%

Step-by-step explanation:

The given random list of numbers from 0 to 4 = 0,1,2,3,4

Therefore, the numbers in the list =5

Also, the number 4 representing a success.

Therefore the number of digits representing success = 1

Now, the probability of a success is given by :-

\text{P(success)}=\dfrac{\text{Number of digits representing }}{\text{Total digits}}\\\\\Rightarrow\text{P(success)}=\dfrac{1}{5}=0.2

In percent , 0.2*100=20\%

Hence, the estimated probability of success = 20%

kupik [55]2 years ago
4 0
It's 20%. idk how to do it, but it says 20%
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Write a vector equation of the line that passes through p(–1 6) and is parallel to a = (3, -1).
Phoenix [80]

Answer:

C

Step-by-step explanation:

(x,y) = p + t(a)

(x,y) = (-1,6) + t(3,-1)

(x,y) - (-1,6) = t(3,-1)

(x + 1, y - 6) = t(3, -1)

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3 years ago
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Morgarella [4.7K]
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Charisse collects antique salt and pepper shakers. The value of her cactus salt and pepper shakers is shown (56.28$). The value
MAVERICK [17]

Answer:

  $53.61

Step-by-step explanation:

It looks like the rate of change is supposed to be -2 2/5% = -2.4% per year. That means the "growth factor" is ...

  growth factor = 1 + growth rate

  growth factor = 1 -2.4% = 1 -0.024 = 0.976

Each year, Charisse's cactus salt and pepper shakers will have their value multiplied by this factor. After 2 years, their value will be ...

  ($56.28)(0.976)(0.976) = $56.28×0.976^2 ≈ $53.61

The value is projected to be $53.61 in 2 years.

6 0
2 years ago
Compute the length of sides AB, AC.
Paha777 [63]
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8 0
2 years ago
Please help me to solve this . Thank you so much .
worty [1.4K]

Answer:

a) N

b) L

c) area I

d) area II

e) area VI

Step-by-step explanation:

a) the points that are 2cm from R are Q, N, M, S. Then, points that are 4cm from P are K, N, R. So, the only one point that works for both is N.

b) the points that are >2cm from R are P, K, L, T. We do not count those are exactly 2cm from R. Then, points that are 4cm from T are R, M, L. Ans is L.

c) <4cm from P, are area I and II. Then area that are >2cm from R are I, VI, and V. So, the only area that works for both is I.

d) <2cm from R, are areas II, III, and IV. Then, <4cm from P, are areas I and II. So, the only one works for both is area II.

e) >4cm from T, are areas I, II, III, VI. Then, >4cm from P, are III, IV, V, VI. Finally, >2cm from R, are areas I, VI, V. The only one that works for all three conditions is area VI.

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