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2 years ago

Please help me with these two questions

Mathematics

2 answers:

In-s [12.5K]2 years ago

7 0

True

-1, -2, -4, -8, ... is the non-arithmetic sequence. It’s geometric as there’s a common ratio of 2.

Vladimir [108]2 years ago

5 0

The median isn’t affected much by one outlier.

sequence: -1,-2,-4,-8 is not arithmetic it is geometric

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Identify the middle line for the function: f (t) = 5sec(3t − π) + 1

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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

<span>(t) = 0 gives
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2 years ago

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2 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:

$Lower =Q_1 - 1.5 * IQR$