75 time because the proabablity of rolling any number each time is 1/10 so 250/10=25 the mutiples of 3 in 1-10 are 3,6,9 so 3*25 =75 tmes
Answer:
c = 32/3
Step-by-step explanation:
Eliminate the fraction by mult. all terms by 8: 96 + 3c = 64.
Subtract 64 from both sides, obtaining: 3c = 32. Solving for c, c = 32/3.
Call (F) the age of the father and (J) the age of Julio
The F & J are related in this way: F=4J
Now you have a restriction in the form of inequality: The sum of both ages has to be greater or equal than 55.
Algebraically that is: F + J ≥ 55
You can substitute F with 4J to find the solution for J:
4J + J ≥ 55
5J ≥ 55
Now divide both sides by 5
5J/5 ≥ 55/5
J ≥ 11
That Imposes a lower boundary for the value of J of 11, meaning that the youngest age of Julio can be 11
Answer:
27 pages.
Step-by-step explanation:
Let l be the number of pages in the long paper and s be the number of pages in the short paper.
We have been given that the total number of pages for both papers is 40. We can represent this information in an equation as:

We are also told that the number of pages in the long paper is one more than two times the number of pages in the short paper. We can represent this information in an equation as:

We will use substitution method to solve system of linear equations.
From equation (1) we will get,

Upon substituting this value in equation (2) we will get,


Upon adding 2l to both sides of our equation we will get,



Let us divide both sides of our equation by 3 we will get,


Therefore, there must be 27 pages in the long paper.