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2 years ago

Easy question, please help

Mathematics

1 answer:

stepladder [879]2 years ago

6 0

Step-by-step explanation:

All you need to do is divide cost by mile since they already factored in the additional $3 for you.

1. 8/5 = $1.60

2. 11/8 = $1.375 = $1.38

3. 23/20 = $1.15

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Muffins are packed and sold in boxes of 4. How many boxes are needed to pack 260 muffins?

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Answer:

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Step-by-step explanation:

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A spinner has four sections. The sections are red, blue, green, and yellow. Andrea spins the spinner many times and records the

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Step-by-step explanation:

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3 years ago

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The diagram shows the dimensions of a new parking lot at Helen's Health Food store. 45 yd g41307181 35 yd The parking lot is squ

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Answer:

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Step-by-step explanation:

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3 years ago

Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their r

olga55 [171]

Answer:

(a):

Dannette Alphonso

$\bar x_D = 4.33$ $\bar x_A = 5.17$

$M_D = 2.5$ $M_A = 5$

$\sigma_D = 3.350$ $\sigma_A = 1.951$

$IQR_D = 7$ $IQR_A = 1.5$

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

$A = \{3,4,4,4,4,5,5,5,5,6,6,11\}$ ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}$

$\bar x_D = \frac{52}{12}$

$\bar x_D = 4.33$ --- Dannette

$\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}$

$\bar x_A = \frac{62}{12}$

$\bar x_A = 5.17$ --- Alphonso

Median

The median is calculated as:

$M = \frac{n + 1}{2}th$

$M = \frac{12 + 1}{2}th$

$M = \frac{13}{2}th$

$M = 6.5th$

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

$M_D = \frac{2+3}{2}$

$M_D = \frac{5}{2}$

$M_D = 2.5$ ---- Dannette

$M_A = \frac{5+5}{2}$

$M_A = \frac{10}{2}$

$M_A = 5$ ---- Alphonso

Standard Deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

So, we have:

$\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}$

$\sigma_D = \sqrt{\frac{134.6668}{12}}$

$\sigma_D = 3.350$ ---- Dannette

$\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}$

$\sigma_A = \sqrt{\frac{45.6668}{12}}$

$\sigma_A = 1.951$ --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

$IQR =Q_3 - Q_1$

Where

$Q_3 \to$ Upper Quartile and $Q_1 \to$ Lower Quartile

$Q_3$ is calculated as:

$Q_3 = \frac{3}{4}*({n + 1})th$

$Q_3 = \frac{3}{4}*(12 + 1})th$

$Q_3 = \frac{3}{4}*13th$

$Q_3 = 9.75th$

This means that $Q_3$ is the mean of the 9th and 7th item. So, we have:

$Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16$ $Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11$

$Q_3 = 8$ ---- Dannette $Q_3 = 5.5$ --- Alphonso

$Q_1$ is calculated as:

$Q_1 = \frac{1}{4}*({n + 1})th$

$Q_1 = \frac{1}{4}*({12 + 1})th$

$Q_1 = \frac{1}{4}*13th$

$Q_1 = 3.25th$

This means that $Q_1$ is the mean of the 3rd and 4th item. So, we have:

$Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2$ $Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8$

$Q_1 = 1$ --- Dannette $Q_1 = 4$ ---- Alphonso

So, the IQR is:

$IQR = Q_3 - Q_1$

$IQR_D = 8 - 1$ $IQR_A = 5.5 - 4$

$IQR_D = 7$ --- Dannette $IQR_A = 1.5$ --- Alphonso

Solving (b): The measures to compare

Measure of center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

$Lower =Q_1 - 1.5 * IQR$

$Upper =Q_3 + 1.5 * IQR$

For Dannette:

$Lower = 1 - 1.5 * 7 = -9.5$

$Upper = 8 + 1.5 * 7 = 18.5$

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

$Valid = 0 \to 18.5$

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

$Lower = 4 - 1.5 * 1.5 =1.75$

$Upper = 5.5 + 1.5 * 1.5 =7.75$

So, the valid data range are:

$Valid = 1.75\to 7.75$

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

4 0

3 years ago

Can you guys please help me? This is due tomorrow, thanks so much

wariber [46]

Answer:

Graph given in Question 18 is a function and the graph given in Question 19 is not a function. Just check the attached figure.

Step-by-step explanation:

From the given graph of a relation, it is easy to establish whether the given relation is a function or not. In order to determine the nature of the relation, there is vertical line test. This test can be done by drawing any vertical line crossing the graph of the given relation. This vertical line has a constant x -a line having constant x.

If this vertical line crosses the graph multiple times, then we can determine that the relation would not be a function. It means the graph contains multiple values of y for a single value of x. This violates the condition of a relation to be a function. Hence, the relation would not be a graph.

But, if the vertical line crosses the graph only one time, then we can determine that the given relation would be a function. It means the graph does not contain multiple values of y for a single value of x.

Solution of the Graph given in Question 18:

From the given graph of a relation, that sounds like a circle. It means the relation shown in current graph is not a function. The reason is simple. If we draw a vertical line, then the line would cross the graph twice. Hence, the given graph of a relation is not a function.

Just check that the value of x=1 will have two y values i.e (1, 5) and (1, -1). Also, at origin, for a value of x = 0, there are multiple values of y. Hence, the graph is not a not a function, but just a relation. Just check the attached figure.

Solution of the Graph given in Question 19:

From the given graph of a relation, it is clear that the graph is a function. The reason is simple. If we draw a vertical line, then the line would cross the graph only one time. Hence, the given graph of a relation is a function.

Just check that the points (0, 0), (2, 3), (-1, 3), (-3, 3) and (1, -3) all make a function as each input value yields an exactly one output function. Hence, the given graph of the relation is a graph. Just check the attached figure.

Keywords: function, relation, ordered pairs, graph, vertical line test

Learn about the graph functions from brainly.com/question/4708956

#learnwithBrainly

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3 years ago