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AleksAgata [21]
1 year ago
12

A party planner is making gift bags for an event. He has 96 pencils,

Mathematics
1 answer:
Yuliya22 [10]1 year ago
5 0

Considering the greatest common divisor, you obtain:

  • the greatest number of gift bags that the party planner can make is 12.
  • the number of each item present in each bag is 8 pencils, 3 erasers and 2 pencil toppers.

<h3>Greatest common divisor </h3>

The greatest common divisor is the largest number that exactly divides two or more numbers at the same time. That is, it is the largest number by which two or more numbers can be divided, resulting in a whole number.

A method to calculate the greatest common factor must follow the following steps:

  • Decompose or separate each number into prime factors.
  • Common factors are noted.
  • In each of the commons, the factor with the smallest exponent is chosen.
  • Multiply the chosen factors.

<h3>This case</h3>

To find the greatest number of gift bags the party planner can make, you find the greatest common divisor.

To do this, decompose the numbers 96, 36 and 24:

  • 96= 2⁵×3
  • 36= 2²×3²
  • 24= 2³×3

The common factors with the smallest exponent are: 2² and 3

So, the greatest common divisor between 96, 26 and 24 is calculated as: 2²×3= 4×3= 12

This means that the greatest number of gift bags that the party planner can make is 12.

To calculate the number of each item present in each bag, you divide the quantity of each item by the quantity of gift bags:

  • 96 pencils÷ 12= 8 pencils
  • 36 erasers÷ 12= 3 erasers
  • 24 pencil toppers÷ 12= 2 pencil toppers

Finally, the number of each item present in each bag is 8 pencils, 3 erasers and 2 pencil toppers.

Learn more about greatest common divisor:

brainly.com/question/11993520

brainly.com/question/18635265

brainly.com/question/6032811

#SPJ1

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Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

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a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

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