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PilotLPTM [1.2K]
3 years ago
7

The ratio of boys to girls in the 6th grade is 2:3 . if there are 24 boys,how many girls are there ?

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

36

Step-by-step explanation:

24/2 and then that times 3

scoundrel [369]3 years ago
5 0

Answer: 36 girls


Step-by-step explanation:

In a ratio, if possible, simplify first. Once the ratio is simplified you can multiply both numbers by the same number and the ratio will remain a ratio (which can be simplified). In this case, we can divide 24 by 2, because there are 24 boys in a ratio of 2 boys to 3 girls. The result is 12, and when we multiply 12 by 3 (the ratio for girls) we get 36. Therefore, the answer is 36 girls.

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Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list
Radda [10]

Answer:

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: Any from 0 to 5.

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

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P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this question:

n = 5, p = 0.3, q = 1 - p = 0.7

So

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

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