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lara31 [8.8K]
4 years ago
13

Caroline works at the movie theater. Her job involves keeping track of the number of matinee and full-price tickets that are sol

d each day. The matinee price is $4.75 per ticket and the full-price ticket is $8.75. Caroline sees that the total ticket sales for the day are $13,325. The sales data also tells her that the number of full-price tickets is 4.5 times the number of matinee price tickets plus 10. PLEASE HELP
Mathematics
1 answer:
professor190 [17]4 years ago
7 0
14.75 is the awnser to the question
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Vera_Pavlovna [14]

Answer:

A) x-3 and D) x+9

Step-by-step explanation:

x^2 + 6x - 27

You need to find two numbers that multiply to -27 and add up to 6

Those two numbers are -3 and 9, because 9 * -3 = -27 and 9+ -3 = 6

So you add those to x and those are your two factored binomials

(x-3) and (x+9)

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3 years ago
Solve for x in the equation 3 x squared minus 18 x + 5 = 47.
ryzh [129]

Answer:

x= 3 +\sqrt{23}, 3 -\sqrt{23}

Step-by-step explanation:

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Then you simplify and get the answers x= 3 +\sqrt{23}, 3 -\sqrt{23}

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15. Work out the<br> area of a<br> rectangle with<br> sides 3.4cm and<br> 4.5cm
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PLEASE HELP WILL MARK BRAINLIEST
Romashka [77]

Answer:

A.The mean would increase.

Step-by-step explanation:

Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.

Presence of outliers effect the measures of central tendency.

The measures of central tendency are mean, median and mode.

The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.

Effect of outliers on mean, median and mode:

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  • Median: The presence of outliers in a data set has a very mild effect on the median of the data.
  • Mode: The presence of outliers does not have any effect on the mode.

The mean of the test scores without the outlier is:

   \bar{x}=\frac{Total of the observations-Outlier value}{n-1} \\=\frac{(86*16)-72}{15} \\=\frac{1304}{15}\\ =86.9333

*Here <em>n</em> is the number of observations.

So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.

The mean increased.

Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.

Conclusion:

After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.

Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.

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