5/6-1/9=?/?-?/? =13/18
1 answer:
15/18 - 2/18 = 13/18
Find the least common multiple of 6 and 9. The LCM is 18. In order to get 18 on the denominator with 6, you have to multiply by 3 because 6 times 3 is 18. You have to multiply 3 to the numerator, so 5 times 3 is 15. Thus, the first fraction is 15/18. To get the second, you need to find what would get you 18 on the denominator with 9. You need 2, so 9 times 2 is 18. You have to multiply 2 to the numerator, so 1 times 2 is 2. Thus, the second fraction is 2/18.
Find the least common multiple of 6 and 9. The LCM is 18. In order to get 18 on the denominator with 6, you have to multiply by 3 because 6 times 3 is 18. You have to multiply 3 to the numerator, so 5 times 3 is 15. Thus, the first fraction is 15/18. To get the second, you need to find what would get you 18 on the denominator with 9. You need 2, so 9 times 2 is 18. You have to multiply 2 to the numerator, so 1 times 2 is 2. Thus, the second fraction is 2/18.
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Binomial
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=10)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (p=5%)4. All trials are random and independent of the others (assumed from context)The number of successes, x, is then given by
where
Substituting values, p=0.05, n=10, X=exactly 1
for X=1 (defective out of n)
P(X=1)=C(10,1)0.05^1*(1-0.05)^(10-1)
=10!/(1!9!)*0.05*0.95^9
=10*0.05*0.0630249
=0.315125 (to 6 places of decimal)
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=10)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (p=5%)4. All trials are random and independent of the others (assumed from context)The number of successes, x, is then given by
Substituting values, p=0.05, n=10, X=exactly 1
for X=1 (defective out of n)
P(X=1)=C(10,1)0.05^1*(1-0.05)^(10-1)
=10!/(1!9!)*0.05*0.95^9
=10*0.05*0.0630249
=0.315125 (to 6 places of decimal)