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bonufazy [111]
3 years ago
11

In a school, 40% of the students have brown eyes. Find the experimental probability that in a group of 4 students, at least one

of them has brown eyes. The problem has been simulated by generating random numbers. The digits 0-9 were used. Let numbers "O", "1", "2", and "3" represent the 40% of students with brown eyes. A sample of 20 random numbers is shown.
7918 7910 2 546 13 90 6075 1230 2 3 86 0793 7 359 3 0 48 2816 6147 5978 5 6 2 1 9 7 3 2 9436 3 806 5971 6173 1 4 3 0
Experimental Probability = [?]% ​
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
5 0

Answer:

70%

Step-by-step explanation:

knowledge

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in sports club there are 800 members . the ratio of soccer members to social members is 5to 3 how many soccer players are there
attashe74 [19]
5:3

5+3=8
8 units

800=8 units
divide oth sides by 8
100=1 units

soccer=5 units=5 times 100=500
social=3 units=3 times 100=300



500 soccer players

5 0
3 years ago
Which of the following numbers has a value between 1.4 and 1 4/5? a. 1.34 b. 170%
Lyrx [107]
Here, numbers are: 1.4 and 1 4/5 = 9//5 = 1.8
Now, you need to know, which number is between 1.4 and 1.8

1.34 is lower than 1.4 So, it can't
170% = 170/100 = 1.7

So, 170% is that number between 1.4 and 1 4/5

Hope this helps!
6 0
3 years ago
Si un kilogramo de arroz cuesta 24.50,cuanto se pagara por 0.75 kg y 4 kg?
klemol [59]

Answer:

0,75kg : 18,38

4kg : 98

Step-by-step explanation:

1) 0,75 x 24,50 = 18,38

2) 4 x 24,50 = 98

Just fix the unit at the end. :)

7 0
3 years ago
A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At
kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

(a) So, the probability of B,

P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

\implies P(A\cap B\cap C)=\frac{1}{8}

7 0
3 years ago
Question 16: Please help, I do not understand this question.
Gala2k [10]

Answer:

A.

Step-by-step explanation:

The lines k l m and n are all  lines of symmetry so the reflections  carry the square onto itself.

7 0
3 years ago
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