I NEED HELP PLEASE, THANKS! :)
1 answer:
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Answer:
Step-by-step explanation:
Hello!
The sample shows the scores for the combined three-part SAT.
Raw data in first attachment.
a.
To arrange the data in a frequency table using class intervals you have to determine the number of intervals you want to use and calculate their width. In this case, the width is given and so is the lower limit of the first interval, you calculate the successive limits by adding the width. The lower limit of the next interval will be the upper limit of the previous one:
1) 800 + 200= 100
First interval [800; 1000)
2) 1000 + 200
Second interval
[1000; 1200)
And so on until you reach the maximum value of the data set,
[1200; 1400)
[1400; 1600)
[1600; 1800)
[1800; 2000)
[2000; 2200)
Then you have to order the data from least to greatest and count how many observations correspond to each value, this way you'll determine the observed frequency for each interval.
Table and histogram in second attachment.
b.
As you can see in the histogram, this distribution is symmetrical centered in the interval [1400; 1600) and there are no outliers observed.
c.
Values around 1400-1600 are the most common ones while scores around 800-1000 or 2000-2200 are more uncommon, in this sample it seems the probability to obtain a perfect score for the combined three-part SAT is extremely low.
I hope you have a nice day!
Answer:
Step-by-step explanation:
either cow or pigs
Either pigs or cow = 40
(1) Let number of pigs be x
then number of cows >twice of pigs = 2x
number of pigs + number of cows = 40
number of cows = 40 -x
40-x > 2x
40 > 3x
x <
x < 12.33
nearest integral number is 12
so its 12
so number of pigs will be 12 so that number of cows = 40 -12 = 28
which is more than twice of number of pigs
28 > 2( 12) = 24
(2) number of pigs + number of cows = 40
for x>12
number of cows = 40 -x
x > 12
-x < -12
adding 40 both sides
40-x < 40-12
number of cows <28