Help please i cant do this and i need help

I NEED HELP FASTT PLEASE

Advocard [28]

It’s the third dot bc X to the tenth power times y divided by three

3 0

3 years ago

Read 2 more answers

The Rogers family drove a total of 482 miles, starting on

Trava [24]

you add up 138+225 and you get 363 then you subtract 482-363 this is how you set it up:

4 8 2

- 3 6 3

------------

you have to borrow because 3 does not go into 2 so you change the 8 into a 7 then you 10 to 2 and you get 12 so this is what it looks like:

4 7 12

- 3 6 3

-------------------

so you then subtract 12-3=9 then 7-6=1 then 4-3=1

so this is what it looks like

4 7 12

--- 3 6 3

--------------------

1 1 9

so on Sunday they rode 119 miles

7 0

2 years ago

A cyclist travels 15km due north and then 18 km due east. How far is she from the starting point?

zavuch27 [327]

Answer:

DONT KNOW SRY

Step-by-step explanation:

8 0

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:

$Lower =Q_1 - 1.5 * IQR$

$Upper =Q_3 + 1.5 * IQR$

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

Wynn is 24 years old and has decided to reduce the number of cups of coffee he buys by 2 cups per day. One cup of coffee typical

Usimov [2.4K]

The total amount he would have at 69 is $343,347.81.

<h3>What is the total amount saved?</h3>

The formula that can be used to determine the future value of the deferred annuity is:

Future value = annuity factor x monthly deposit

Annuity factor = {[(1+r)^n] - 1} / r

Where:

r = interest rate = 5.5 /2 = 2.75%
n = number of payments = 2 x (69 - 24) = 90

Amount he would save every 6 months:

amount saved per day = $2.50 x 2 = $5
Amount saved per month : $5 x 30 = $150
Amount saved every 6 months = $150 x 6 = $900

Future value : 900 x {[(1.0275^90) - 1] / 0.275}= $343,347.81

To learn more about annuities, please check: brainly.com/question/24108530

#SPJ1

5 0

1 year ago