pretty much about the same as before.
a = weight of a large box
b = weight of a small box.
we know their combined weight is 65 lbs, thus a + b = 65.
we also know that the truck has 60 large ones, and 55 small ones, thus 60*a is the total weight for the large ones and 55*b is the total weight for the small ones, and we know that is a total of 3775, 60a + 55b = 3775.
Answer:
1/6
Step-by-step explanation:
1/2/3 is a sixth of an apple that all 3 people equally shared.
Main street is vertical to park entrance
Answer:
B
Step-by-step explanation:
Data set D does not contain the value 128, which is the median value.
Data set C does not contain the outlier value 91.
Data set A contains value 168, which does not show up on the plot.
The only remaining choice is B.
_____
In order, the data values of set B are ...
... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152
The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)
Answer:
4(2e - 3)(3e + 1)
Step-by-step explanation:
Given
24e² - 28e - 12 ← factor out 4 from each term
= 4(6e² - 7e - 3) ← factor the quadratic
Consider the factors of the product of the e² term and the constant term which sum to give the coefficient of the e- term.
product = 6 × - 3 = - 18 and sum = - 7
The factors are - 9 and + 2
Use these factors to split the e- term
6e² - 9e + 2e - 3 ( factor the first/second and third/fourth terms )
= 3e(2e - 3) + 1 (2e - 3) ← factor out (2e - 3) from each term
= (2e - 3)(3e + 1)
Then
24e² - 28e - 12 = 4(2e - 3)(3e + 1) ← in factored form
