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4 years ago

What is 7/11 in simplest form?

1 answer:

3 0

7/11 is already simplified, it cannot be simplified any further as both the numerator and denominator are prime numbers and it is not an improper fraction

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Explain how to divide by to use the terms dividend divisor and quotient

Snowcat [4.5K]

Answer:

The number which is divided is called the dividend. The number which divides is called the divisor. The number which is the result of the division is called the quotient. If there is any number left over, it is called the remainder.

Step-by-step explanation:

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3 years ago

Number 7 help ASAP ASAP

MariettaO [177]

Answer:

Step-by-step explanation:

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3 years ago

Simplify this expression 5a/11a

barxatty [35]

You could cross cancel the a and your final answer would be 5/11

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4 years ago

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A box at the post office has dimensions as shown below. A rectangular prism with a length of 12 inches, width of 8 inches, and h

steposvetlana [31]

Answer: 1536 inches

Step-by-step explanation:

Fron the question, a rectangular prism has a length of 12 inches, width of 8 inches, and then a height of 16 inches. The volume can be gotten by multiplying the length by the width and height.

Volume= length × width × height

Volume = 12 × 8 × 16

Volume = 1536 inches

The volume of the box is 1536 inches.

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3 years ago

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

Scilla [17]

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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3 years ago