Given,
The sum of three integers is 92
So,
Let,
The first integer be "x"
The second integer be "y"
The third integer be "z"
Now,
According to the question,
y = 3x ..............equation (1)
z = 2x - 10 .............. equation (2)
x + y + z = 92 ..............equation (3)
Now,
Substituting the value of "y" and "z" from equation (1) and (2), we get,
x + (3x) + (2x - 10) = 92
x + 3x + 2x - 10 = 92
6x - 10 = 92
6x = 92 + 10
x = 102 / 6
x = 17
Now,
substituting the value of "x" in equation (1)
y = 3 (17)
y = 51
Now,
Substituting the value of "x" in equation (2),
z = 2 (17 ) - 10
z = 34 - 10
z = 24
So, the numbers are 17, 51 and 24
NEVER HATE MATH!!!
There are 4.5 ounces of applesauce in each container
Answer:
Step-by-step explanation:
The problem relates to filling 8 vacant positions by either 0 or 1
each position can be filled by 2 ways so no of permutation
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
b )
Probability of opening of lock in first arbitrary attempt
= 1 / 256
c ) If first fails , there are remaining 255 permutations , so
probability of opening the lock in second arbitrary attempt
= 1 / 255 .
Answer:
m puq
Step-by-step explanation:
divide both sides by a
subtract b from both sides
multiply all terms on both sides by -1
c=b-d/a
