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polet [3.4K]
3 years ago
8

Add and simplify: 9/16 + 1/2=? 5/8 7/16 1 1/16 1 5/9

Mathematics
2 answers:
Rina8888 [55]3 years ago
5 0

Answer:

1 1/+6

Step-by-step explanation:

9/16 + 8/16=17/16=1 1/16

Dovator [93]3 years ago
3 0

Answer:

1 1/16.

Step-by-step explanation:

The Lowest Common Denominator  of 2 and 16 is 16 so we have:

9/16 + 1/2

= 9/16 + 8/16

= 17/16

= 1 1/16.

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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What
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Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

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Answer:

Step-by-step explanation:

r=4/(-2)=-2

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Step-by-step explaidknation:

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