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3 years ago

What is the answer to this question

Mathematics

1 answer:

nikdorinn [45]3 years ago

3 0

The answer would be c because when you find what the variable is and then you would add g to 33 to try to get 48

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Find the lengths of DE and EF. Note: Segments BC and DF are parallel.

balandron [24]

c-de=9cm; ef=15cm hope this helps

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3 years ago

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HELP URGENTLY:

AfilCa [17]

Answer:

option C

Step-by-step explanation:

the solutions to function x^2 -9x +14 are (x-7) and (x-2)

through the zero product property

(x-7)=0

x=7

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4 years ago

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Speedy Swift is a package delivery service that serves the greater Atlanta, Georgia, metropolitan area. To maintain customer loy

Paladinen [302]

Answer:

See Explanation and attachments

Step-by-step explanation:

Given

See question for data

Solving (a):

The scale used is an ordinal scale.

Ordinals scale uses hierarchy arrangement and the data can be arranged as:

$EARLY ->> ON-TIME ->> LATE ->> LOST$

The variable of the delivery performance is qualitative because it is non-numerical

Solving (b): Frequency table

From the table, we have:

$On - Time = 57$ $Early = 19$ $Late = 9$ $Lost = 2$

So, the frequency distribution table is:

$\begin{array}{cc}{Performance}&{Frequency}&{On-Time}&{57}&{Early}&{19}&{Late}&{9} & {Loss} & {2} & {Total} & {87}\end{array}$

Solving (c): Frequency table.

To do this, we add another column (Relative Frequency) to the above table.

The relative frequency is calculated as:

$Relative\ Frequency = \frac{Frequency}{Total}$

So, we have:

$On - Time = \frac{57}{87} = 0.656$ $Early = \frac{19}{87} = 0.218$

$Late = \frac{9}{87} = 0.103$ $Lost = \frac{2}{87} = 0.023$

So, the frequency distribution table is:

$\begin{array}{ccc}{Performance}&{Frequency}&{Relative\ Frequency} & {On-Time}&{57}&{0.656}&{Early}&{19}&{0.218} & {Late}&{9} & {0.103} & {Loss} & {2} & {0.023} & {Total} & {87}&{1}\end{array}$

Solving (d & e): See attachment 1 for bar chart & attachment 2 for pie chart

Solving (f):

From the question, we understand that the object is to return 99% early or on time and never to lose a package.

The analysis is as follows:

Early and On-Time (ET) packages

$ET = \frac{Early + On-Time}{Total} * 100\%$

$ET = \frac{57 + 19}{87} * 100\%$

$ET = \frac{76}{87} * 100\%$

$ET = \frac{7600}{87} \%$

$ET = 87.4 \%$

Lost packages

$Lost = \frac{Lost}{Total} * 100\%$

$Lost = \frac{2}{87} * 100\%$

$Lost = \frac{200}{87} \%$

$Lost = 2.30\%$

From the above analysis, we can see that 87.4% of the packages were delivered early enough and 2.30% were lost.

The fraction of packages delivered early can be improved and the fraction of lost packages can be reduced by exploring the chances of taking alternative routes when possible.

3 0

3 years ago

0-4, 5-9, 10-14, 15-19, 20-24 Create a frequency distribution

timofeeve [1]

Answer:

the common difference is 4

Step-by-step explanation:

we can normally find out that the difference between the first item and the second item is 4

and the difference between the third item and fourth item is 4

3 0

4 years ago

Ocelyn Zamora

mamaluj [8]

Answer:

Yuhh the answers c dawhggg trus me i just did the same thing lawl!

Step-by-step explanation:

5 0

4 years ago