Answer:
Step-by-step explanation:
Given
The above data
Required
The class width
To do this, we simply calculate the difference between the class limits of any one of the classes.
Taking 1187–1246 as a point of reference, the class width is:
Step-by-step explanation:
Using the identity a² - b² = (a + b) (a - b)
=> (3 {p - q})² - (5 {q - r})²
=> [(3 {p - q}) + (5 {q - r})] [(3 {p - q}) - (5 {q - r})]
=> [(3p - 3q) + (5q - 5r)] [(3p - 3q) - (5q - 5r)]
=> [3p - 3q + 5q - 5r] [3p - 3q - 5q + 5r]
=> [3p + 2q - 5r] [3p - 8q + 5r]
I remember foing this i think it is x>453
I am not 100 percebt sure
Answer:
308, 398, 421, 433, 570
Step-by-step explanation:
To start we know the smallest is 308 cans, so that is the first class.
Next, it says the range is 262. The range is maximum minus the minimum. Since we know the minimum(308), that means 308 + 262 = 570 and 570 is the maximum or the class with most collected cans.
Then, it says the median or the middle number is 421, the middle number in a class is 3. So, the third class collected a total of 421 cans.
After that, it says the two classes in the three hundred, or the first and second class, has a total of 706 cans. So, if we do 706 - 308 we get the second class. Which collected 398 cans.
Finally, it says the total all the class collected is 2130. So, if we add all the values we have know and subtract it from the total of all the class we get the missing class or the fourth class. So, 308 + 398 + 421 + 570 = 1697, and 2130 -1697 = 433. Which the fourth class.
Least to greatest:
308
398
421
433
570
