First let's write out the inequality before choosing a graph.
x apples each weighing 1/3 of a pound: 1/3x
y pounds of grapes: y
So...
1/3x + y < 5
The maximum weight is 4 pounds since the total weight of both the grapes and apples are less than 5.
In the y-axis, the first, third, and fourth graphs already exceed the capacity of 5 pounds.
So, by process of elimination, the correct graph for this problem is the second one.
Answer:
The number of ways is equal to 
Step-by-step explanation:
The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are
n1 x n2 x ... x ni
Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.
For example : We want to put 4 different objects in a row.
The total ways to do this are
ways.
Using the multiplication principle and the n-factorial number :
The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ 
Answer:
1/5
Step-by-step explanation:
i had a similar question
For each roll you start with paying 2 dollars and you only with 10 dollars one out of 6 rolls (on average).
So the cost for one play is 2 dollars and your win is 10/6.
Value is -2+10/6=-1/3 dollars
So you lose 1/3 dollars on average with each game
since you have no limited rolls u put 1/5
this from another question but both same just different numbers
Answer: 1
Step-by-step explanation:
Change in y/change in x
starting at y intercept. up 1 unit (change in y) and over 1 unit (change in x) to get to next point on line.
slope is 1/1 which reduces to 1.
To solve this problem, let us first lay out all the
factors of each number.
48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
56 : 1, 2, 4, 7, 8, 14, 28, 56
The greatest number of bouquets that can be made would be
equal to the greatest common factor of the two numbers. In this case it would
be 8.
Answer:
<span>8 bouquets</span>