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Anit [1.1K]
2 years ago
14

Wholesome Pizza surveyed a random sample of 404040 Trinity Food Festival attendees about their favorite type of pizza. Of the at

tendees surveyed, 212121 said that sausage pizza was their favorite type of pizza. There are 160016001600 Trinity Food Festival attendees. Based on the data, what is the most reasonable estimate for the number of Trinity Food Festival attendees whose favorite type of pizza is sausage pizza?
Mathematics
2 answers:
ale4655 [162]2 years ago
6 0
For this case the first thing we should do is find the percentage of people surveyed who said that pizza pizza was their favorite type of pizza.
 We have then:
 x = (21/40) * (100)

x = 52.5%
 We are now looking for 52.5% of 1600 people.
 For this, we make the following rule of three:
 1600 ----------> 100%
 x ---------------> 52.5%
 Clearing the value of x we have:
 x = (52.5 / 100) * (1600)

x = 840
 Answer:
 
the most reasonable estimate for the number of Trinity Food Festival attendees whose favorite type of pizza is sausage pizza is:
 
x = 840
crimeas [40]2 years ago
4 0

Answer: The answer is 840

Step-by-step explanation:

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Hello,

If b represents any real number,

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2.4 Backgammon. Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two di
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Answer:

The claim is not fair because:

  • Probability of roliing two 6s = Probability of rolling two 3s = 1/36

Step-by-step explanation:

The <em>probaility</em> of an event is defined as the number of favorable outcomes divided by the number of total possible events.

P (event E) = number of outcomes for event E / number of possible events

1. <u>As </u><u>first step</u><u>, you may draw a table to find the </u><u>sample space</u><u> (set of all possible outcomes)</u>.

<u>Sample space</u>

The results of the rolling two dice are summarized in this table:

                    Second roll    1       2       3       4        5      6

First roll

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

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

  3                                     (3,1)  (3,2)  <u>(3,3)</u>   (3,4)  (3,5)  (3,6)

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

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

  6                                     (6,1)  (6,2)  (6,3)  (6,4)  (6,5)  <u>(6,6)</u>

<u></u>

2.<u>Now, in that table, you can observe</u>:

The results (3,3) and (6,6) are highlited.

a) Total number of events: 6 × 6 = 36

b) Nnmber of outcomes for the event rolling two 6s (6,6): 1

c) Number of outcomes for the event rolling two 3s (3,3): 1

3) <u>Next, you can calculate the probabilities:</u>

a) Probability rolling two 6s = 1 / 36

b) Probability of rolling two 3s: 1 / 36

4. <u>Conclusion</u>: the probabilities prove that rolling two 6s is just as likely as rolling two 3s.

7 0
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