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lbvjy [14]
2 years ago
11

c) There are 32 students in a class. If the ratio of the number of boys and girls is 5:3, find the number of boys and girls. ​

Mathematics
2 answers:
stepladder [879]2 years ago
5 0

Answer:

The no. of boys = 20

The no. of girls = 12

Step-by-step explanation:

Ratio of no. of boys to girls = 5 : 3

[5 multiplied with something, and 3 multiplied with something, added together makes 32.Thus, that something = x]

5x + 3x = 32

= 8x = 32

=x = 32÷8 = 4

5x = 5 x 4 = 20

3x = 3 x 4 = 12

Nookie1986 [14]2 years ago
3 0

boys : girls = 5 : 3 (total ratio = 8)

boys = 5/8 x 32 = 20

girls = 3/8 x 32 = 12

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Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a
lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

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3 years ago
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