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Roman55 [17]
3 years ago
11

You have 125% of the tickets required for a souvenir what fraction of the required tickets do you have do you have do you need m

ore tickets for the souvenir
Mathematics
2 answers:
Mademuasel [1]3 years ago
6 0

Answer:

1)

Fraction is:

\dfrac{5}{4}

2)

No, we will not need any more tickets.

Step-by-step explanation:

It is given that:

You have 125% of the tickets required for a souvenir.

We are asked to find:

what fraction of the required tickets do you have?

do you have do you need more tickets for the souvenir?

1)

Firstly we have to find the fraction of the tickets.

i.e. we have to represent 125% in term of the fraction.

We know that 125% could be represented as:

\dfrac{125}{100}=\dfrac{25\times 5}{25\times 4}=\dfrac{5}{4}

Hence, \dfrac{5}{4} is the required fraction.

2)

No, we will not  need any more tickets for the souvenir.

Because you will actually have 1/4 of the tickets needed left over. ( Also it could be seen that the fraction is greater than 1).

drek231 [11]3 years ago
3 0
You have 5/4 of the required tickets

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Answer:

a) P(Y > 76) = 0.0122

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Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

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              (b) If two men are chosen at random from the population, find

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Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{S.D}) > \frac{( 76- mean)}{S.D})

                 = P(Z >  \frac{( 76- mean)}{S.D})

                 = P(Z > \frac{76 - 69.7}{2.8})

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

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⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

\frac{S.D}{\sqrt{N} } = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } })) > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- 69.7)}{1.979899 }))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007

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