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IgorLugansk [536]
4 years ago
7

Five men can install 200 yards of pipeline in an eight hour day three men are added to the job assuming the individuals rates re

main the same how many days will it take the entire crew to install 2240 yards of pipeline
Mathematics
1 answer:
Novay_Z [31]4 years ago
3 0

Answer:

7 days

Step-by-step explanation:

Five men do 200 yards in one day

One man does 200/5 = 40 yards in 1 day.

=============

Now you want to know something about 8 men

8 men can do 40 * 8 = 320 yards in 1 day

=============

2240 yards / 320 yards = 7 days

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For which table would 70 be the missing value,x?
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data test;

input landval improval totval salepric saltoapr city $6. season $8.;

datalines;

  30000     64831     94831    118500   1.25  A    spring

  30000     50765     80765     93900    .         winter

  46651     18573     65224         .   1.16  B      

  45990     91402         .    184000   1.34  C    winter

  42394         .     40575    168000   1.43        

      .      3351     51102    169000   1.12  D    winter

  63596      2182     65778         .   1.26  E    spring

  56658     53806     10464    255000   1.21      

  51428     72451         .         .   1.18  F    spring

  93200         .      4321    422000   1.04      

  76125     78172     54297    290000   1.14  G    winter

      .     61934     16294    237000   1.10  H    spring

  65376     34458         .    286500   1.43       winter

  42400         .     57446         .    .    K    

  40800     92606     33406    168000   1.26  S    

;

run;

1. Number of missing values vs. number of  non missing values in each variable

The first thing we are going to look at the variables that have a lot of missing values. For numerical variables, we use proc means with the options n and nmiss.

proc means data = test n nmiss;

 var _numeric_;

run;

Variable N N Miss

landval 13 2

improval 12 3

totval 12 3

salepric 11 4

saltoapr 13 2

For character variables, we can use proc freq to display the number of missing values in each variable.

proc freq data = test;

 tables city season ;

run;

 

city Frequency Percent Cumulative Cumulative

Frequency Percent

A 1 10 1 10

B 1 10 2 20

C 1 10 3 30

D 1 10 4 40

E 1 10 5 50

F 1 10 6 60

G 1 10 7 70

H 1 10 8 80

K 1 10 9 90

S 1 10 10 100

Frequency Missing = 5

season Frequency Percent Cumulative Cumulative

Frequency Percent

spring 4 44.44 4 44.44

winter 5 55.56 9 100

Frequency Missing = 6

2. Number of missing values in each observation

We can also look at the number of missing values in each observation. For example, we can use SAS function cmiss to store the number of missing values from both numeric and character variables in each observation.

data test1;

 set test;

 miss_n = cmiss(of landval -- season);

run;

proc print data = test1;  

run;

                                                                 

Obs landval improval totval salepric saltoapr city season miss_n

1 30000 64831 94831 118500 1.25 A spring 0

2 30000 50765 80765 93900 .  winter 2

3 46651 18573 65224 . 1.16 B  2

4 45990 91402 . 184000 1.34 C winter 1

5 42394 . 40575 168000 1.43   3

6 . 3351 51102 169000 1.12 D winter 1

7 63596 2182 65778 . 1.26 E spring 1

8 56658 53806 10464 255000 1.21   2

9 51428 72451 . . 1.18 F spring 2

10 93200 . 4321 422000 1.04   3

11 76125 78172 54297 290000 1.14 G winter 0

12 . 61934 16294 237000 1.1 H spring 1

13 65376 34458 . 286500 1.43  winter 2

14 42400 . 57446 . . K  4

15 40800 92606 33406 168000 1.26 S  1

3. Distribution of missing values

We can also look at the patterns of missing values. By default the MI procedure will output missing data patterns for the variables in the specified datasets. If no var statement is specified Proc MI will output a table for the all the variables in a dataset. The ods select statement tells SAS to only output the "Missing Data Patterns" table.

proc mi data=test;

ods select misspattern;

run;

Missing Data Patterns

Group landval improval totval salepric saltoapr Freq Percent Group Means

landval improval totval salepric saltoapr

1 X X X X X 4 26.67 50896 72354 48250 207875 1.215

2 X X X X . 1 6.67 30000 50765 80765 93900 .

3 X X X . X 2 13.33 55124 10378 65501 . 1.21

4 X X . X X 2 13.33 55683 62930 . 235250 1.385

5 X X . . X 1 6.67 51428 72451 . . 1.18

6 X . X X X 2 13.33 67797 . 22448 295000 1.235

7 X . X . . 1 6.67 42400 . 57446 . .

8 . X X X X 2 13.33 . 32643 33698 203000 1.11

You will notice that this report only contains information for numeric variables not character. Another approach to achieve the same output could be via making use of formats, which then allows for character variables to be included. The order of the patterns is different but the information is still the same.

proc format;

value nm . = '.' other = 'X';

value $ch ' ' = '.'other = 'X';

run;

proc freq data=test;

table landval*improval*totval*salepric*saltoapr*city*season / list missing nocum;

format _numeric_ nm. _character_ $ch.;

run;

landval improval totval salepric saltoapr city season Frequency Percent

. X X X X X X 2 13.33

X . X . . X . 1 6.67

X . X X X . . 2 13.33

X X . . X X X 1 6.67

X X . X X . X 1 6.67

X X . X X X X 1 6.67

X X X . X X . 1 6.67

X X X . X X X 1 6.67

X X X X . . X 1 6.67

X X X X X . . 1 6.67

X X X X X X . 1 6.67

X X X X X X X 2

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