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PIT_PIT [208]
2 years ago
7

Six cups of coffee cost $9. Write an equation to represent this statement. Solve the equation to find the cost of one cup of cof

fee.
Mathematics
2 answers:
guapka [62]2 years ago
8 0

Answer:

c= 3/2 or 1.5$

Step-by-step explanation:

c is for the cost of one cup of coffee

6c = 9$

c= 9/6

c= 3/2 or 1.5

Alex17521 [72]2 years ago
4 0
<h3><u>Question</u><u>:</u><u>-</u></h3>

Six cups of coffee cost $9. Write an equation to represent this statement. Solve the equation to find the cost of one cup of coffee.

<h3><u>Answer</u><u>:</u><u>-</u></h3>

Each cup of coffee costs $1.5

<h3><u>Explanation</u><u>:</u><u>-</u></h3>

Let the cost of one cup of coffee be x.

=> 6x = 9

=> x = 9/6

=> x = 1.5

Cost of one cup of coffee = $1.5

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Ksenya-84 [330]
Decreased as getting more for less/same amount makes it so the worth of euro has went down
8 0
3 years ago
Find the greatest common factor for the group of numbers. 12 ​, 28 ​, 24
irakobra [83]
The greatest common factor, or GCF, is quite self-explanatory. Between two numbers, you have to find the greatest number that can be divided by both numbers without any remainder. This is useful in solving addition and subtraction operations involving fractions. The technique to do this is to place the numbers to the right. Then, place the factors to the left. I'm gonna show the solution so that you can understand better.

  4   |   12    28    24
       ----------------------
       |     3     7      6
       ----------------------

Since all numbers can be divided by 4, you place it on the left side. Then, place the quotients on the next row below. Since, there is no more common factor, it ends with the 2nd row. Then, multiply all the left numbers with the numbers in the very last column.

GCF = 4×3×7×6
GCF = 504
8 0
3 years ago
Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,
Bas_tet [7]

Answer:

a) \bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

Median = 25

b) Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) Midrange = \frac{70+13}{3}=41.5

d) Q_1 = \frac{20+21}{2} =20.5

Q_3 =\frac{35+35}{2}=35

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

Median = 25

Part b

The mode is the most repeated value on the dataset on this case would be:

Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

Midrange = \frac{Max+Min}{2}

And if we replace we got:

Midrange = \frac{70+13}{3}=41.5

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

Q_1 = \frac{20+21}{2} =20.5

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

Q_3 =\frac{35+35}{2}=35

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

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3 years ago
Walker is reading a book that is 792 pages. He reads 15 pages a day during the week, and 25 pages a day during the weekend. Afte
liubo4ka [24]

Answer:

167 Pages Left To Read

Step-by-step explanation:

15*5=75

75*5=375

375+ 25*10=625

792-625=167

8 0
3 years ago
What is the discriminant of -2x to the power of 2 -x-1=0
Lorico [155]
Uhhh 67904- 67903-iwbisbeus e
8 0
2 years ago
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