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cricket20 [7]
3 years ago
6

Suppose two dice are tossed and the numbers on the upper faces are observed. let s denote the set of all possible pairs that can

be observed. [these pairs can be listed, for example, by letting (2, 3) denote that a 2 was observed on the first die and a 3 on the second.]
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Suppose two dice are tossed and the numbers on the upper faces are observed. let s denote the set of all possible pairs that can be observed.

Answer: If two dice are tossed, there are 6^{2} =36 possible outcomes.The possible pairs that can be observed are given in a set denoted by s below:

\left[\begin{array}{cccccc}(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)\end{array}\right]

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Answer:

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Step-by-step explanation:

The total you are buying is 112 pounds.  The can you have picked up is 34 pounds; this leaves

112-34 = 78 pounds remaining.

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Sarah bought 7 more songs than Betsy.If Sarah bought 21 songs from Apple Music, how many songs did Betsy buy?
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A rectangular auditorium seats 1564 people. the number of seats in each row exceeds the number of rows by 1212. find the number
yanalaym [24]
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4 0
3 years ago
The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is
marysya [2.9K]

Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.

It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})

                   =P(Z>3)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

P(X

                   =P(Z

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

P(1400

                              =P(-2

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

P(1500

                              =P(-1

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

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