Answer:
3/2
Step-by-step explanation:
My reasoning is that the sum of the numbers in the left is equal to the sum of the numbers in the right.
Therefore 6=2y+3
2y=6-3
2y=3
y=1.5 or 1and a half which is equivalent to 3/2
1.) Number of Total Shirts = 220
In Figure 1, Red boxes represent "<span>short sleeve shirts" which is 7 in number whereas Blue boxes represent "long sleeve shirts".
1 box contains = 220/10 = 22 shirts
Number of short sleeves = 22 * 7 = 154
<span>
In short, Your Final Answer would be: 154 </span></span>2.) Number of Total Subscribers = 260
In Figure 1, Red boxes represent "<span>regular subscribers" which is 10 in number whereas Blue boxes represent "premium subscribers".
1 box contains = 260/13 = 20 subscribers.
Regular subscribers = 20 * 10 = 200
<span>
In short, Your Final Answer would be: 200</span></span>
3.) Number of Total People = 360
In Figure 1, Red boxes represent "<span>Males" which is 3 in number whereas Blue boxes represent "females".
1 box contains = 360/12 = 30 people
Number of males: = 30 * 3 = 90
<span>
In short, Your Final Answer would be: 90
Hope this helps!</span></span>
Let d represent number of days and n represent number of workers.
We have been given that when building a house, the number of days required to build varies inversely with with the number of workers.
We know that the equation
represents the relation where y is inversely proportional to x and k is the constant of proportionality.
So our required equation would be 
Upon substituting our given values, we will get:



Since constant of proportionality is 665, so our equation would be
.
To find the number of days it will take to build a similar house with 5 workers, we will substitute
in our equation as:


Therefore, it will take 133 days for 5 workers to build a similar house.
Consider what the slope of this line would be. Slope is rise/run; this line rises 7 (5 - (-2)) and runs 0 (0 - 0). This means that the slope would be 7/0. Dividing by zero is not possible, therefore, it cannot be written in slope-intercept form.