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Alekssandra [29.7K]
3 years ago
7

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam

e-day ticket is . For one performance, advance tickets and same-day tickets were sold. The total amount paid for the tickets was . What was the price of each kind of ticket
Mathematics
1 answer:
fomenos3 years ago
3 0

Answer:

Advance tickets=$25

Same-day tickets=$15

Step-by-step explanation:

Complete question below:

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $ 40. For one performance, 25 advance tickets and 30 same-day tickets were sold. The total amount paid for the tickets was $1075 . What was the price of each kind of ticket?

Let

advance tickets=x

Same-day tickets=y

Combined cost of advance and same-day tickets=$40

It means,

x+y=40 Equ (1)

25 advance tickets and 30 same-day tickets=$1075

It means,

25x+30y=1075 Equ(2)

From (1)

x+y=40

x=40-y

Substitute x=40-y into (2)

25x+30y=1075

25(40-y)+30y=1075

1000-25y+30y=1075

5y=1075-1000

5y=75

Divide both sides by 5

5y/5=75/5

y=15

Recall,

x+y=40

x+15=40

x=40-15

=25

x=25

Advance tickets=$25

Same-day tickets=$15

Check

25x+30y=1075

25(25)+30(15)=1075

625+450=1075

1075=1075

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8 0
2 years ago
3. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population p
damaskus [11]

Answer:

A)sample proportion = 0.17,  the sampling distribution of p can be calculated/approximated with normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

B) 0.869

C)0.9668

Step-by-step explanation:

A) p ( proportion of population that spends more than $100 per week) = 0.17

sample size (n)= 800

the sample proportion of p = 0.17

standard error of p = \sqrt{\frac{p(1-p)}{n} } = 0.013281

the sampling distribution of p can be calculated/approximated with

normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

B) probability that the sample proportion will be +-0.02 of the population proportion

= p (0.17 - 0.02 ≤ P ≤ 0.17 + 0.02 ) = p( 0.15 ≤ P ≤ 0.19)

z value corresponding to P

Z = \frac{P - p}{standard deviation}

at P = 0.15

Z =  (0.15 - 0.17) / 0.013281 = = -1.51

at P = 0.19

z = ( 0.19 - 0.17) / 0.013281 = 1.51

therefore the required probability will be

p( -1.5 ≤ z ≤ 1.5 ) = p(z ≤ 1.51 ) - p(z ≤ -1.51 )

                           = 0.9345 - 0.0655 = 0.869

C) for a sample (n ) = 1600

standard deviation/ error = 0.009391 (applying the equation for calculating standard error as seen in part A above)

therefore the required probability after applying

z = \frac{P-p}{standard deviation} at p = 0.15 and p = 0.19

p ( -2.13 ≤ z ≤ 2.13 ) = p( z ≤ 2.13 ) - p( z ≤ -2.13 )

                               = 0.9834 - 0.0166 = 0.9668

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3 years ago
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