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Cloud [144]
3 years ago
12

The word "less than" represent what operation?

Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

  • The term 'less than' is used to indicate the relationship between two values.
  • The term 'less than' symbolically represented as '<'.

Step-by-step explanation:

  • The term 'less than' is used to indicate the relationship between two values.

It expresses that one value will be lesser than the other.

The term 'less than' symbolically represented as '<'.

We can implement certain operations like comparing values, heights, and numbers.

For example, there are 3 pencils in one bag A, and 8 pencils in bag B.

When we compare the two, we can easily observe that bag A has fewer pencils than bag B.

Thus,

Pencils in Bag A       <         Pencils in Bag A

           3                     <                    8

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Lizzie rolls two dice. What is the probability that the sum of the dice is:
zhenek [66]

Answer:

A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A)  = \dfrac{12}{36} = \dfrac{1}{3}

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)}  Count 1

Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

P(B)  = \dfrac{15}{36} = \dfrac{5}{12}

C. Divisible by 3 and less than 7:

P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

P(A \cap B) = \dfrac{7}{\text{36}}

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Setler [38]

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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