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elena-14-01-66 [18.8K]
2 years ago
14

a charitable organization in Lanberry is hosting a black tie benefit. Yesterday, the organization sold 55 regular tickets and 49

VIP tickets, raising $9,663.

Mathematics
1 answer:
jenyasd209 [6]2 years ago
7 0

Answer:

  • regular: $67
  • VIP: $122

Step-by-step explanation:

Let r and v represent the costs of a regular and VIP ticket, respectively. The two sales can be represented by ...

  55r +49v = 9663

  79r +84v = 15541

We can subtract 7 times the second equation from 12 times the first to eliminate the v terms.

  12(55r +49v) -7(79r +84v) = 12(9663) -7(15541)

  660r +588v -553r -588v = 115,956 -108,787 . . . . . eliminate parentheses

  107r = 7169 . . . . . . . simplify

  r = 67 . . . . . . . . . . . divide by 107

Using this value in the first equation, we have ...

  55(67) +49v = 9663

  v = 5978/49 . . . . . . . . . subtract 3685, divide by 49

  v = 122

A regular ticket costs $67; a VIP ticket costs $122.

_____

<em>Additional comment</em>

When using "elimination" to solve a system of equations, you're looking for coefficients of the same variable that are related by small factors. Preferably, one coefficient is the same as, or a small multiple of, the other. Here, 55 and 79 (the coefficients of x) are not related by an integer, or a couple of small integers. On the other hand, the y-coefficients 49 and 84 have a common factor of 7, and are in the ratio 7:12, a pair of small numbers. This is why we chose to eliminate the y-variable.

The x-variable could be eliminated using 55 and -79 as multipliers of the equations. This results in larger numbers, and more chance for error. (Errors tend to creep in when computing or copying large numbers.)

Of course, any of several machine methods could be used to solve these equations, including graphing and matrix-solving functions. Here, we tried to honor the requirement to solve by elimination.

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Answer:

Interquartile range => 7

Third quartile => 22

Range => 13

First quartile => 15

Step-by-step explanation:

Order your data set, from the least amount to the highest amount:

$12, $14, $15, $15, $15, $15, $16, $22, $24, $25

Interquartile range (IQR) = third quartile (Q3) - first quartile (Q1)

Q1 = the middle value of the lower part of the data set, from the median to your left.

Q3 = the middle value of the upper part of the data set, from the median to your right.

The median lies between the 5th and 6th value that is enclosed in the parenthesis below:

$12, $14, ($15), $15, $15,[Median], $15, $16, ($22), $24, $25

The median divides the data set into upper and lower part.

Median = \frac{15 + 15}{2} = 15

First quartile: Q1 = $15

Third quartile: Q3 = $22

IQR = $22 - $15 = $7

Range = highest amount - least amount = 25 - 12 = $13

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Step-by-step explanation:

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3 years ago
The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
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Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

                      Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

                      Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

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Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

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