Answer:
Step-by-step explanation:
Our inequality is |125-u| ≤ 30. Let's separate this into two. Assuming that (125-u) is positive, we have 125-u ≤ 30, and if we assume that it's negative, we'd have -(125-u)≤30, or u-125≤30.
Therefore, we now have two inequalities to solve for:
125-u ≤ 30
u-125≤30
For the first one, we can subtract 125 and add u to both sides, resulting in
0 ≤ u-95, or 95≤u. Therefore, that is our first inequality.
The second one can be figured out by adding 125 to both sides, so u ≤ 155.
Remember that we took these two inequalities from an absolute value -- as a result, they BOTH must be true in order for the original inequality to be true. Therefore,
u ≥ 95
and
u ≤ 155
combine to be
95 ≤ u ≤ 155, or the 4th option
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
Given :
Cost of single bottle, C = $1.69 .
Cost of 4 bottles, B = $5 .
You bought one bottle last week for $1.69 and one bottle this week at the 4 for $5 price.
To Find :
Did you pay more or less for the bottle this week at 4 for $5? How much more or less?
Solution :
Effective price per bottle if we purchase 4 is, E = 5/4 = $1.25 .
Regular price, R = $1.69 .
Change in price, C = $( 1.69 - 1.25 ) = $0.44 .
Therefore, we are paying $0.44 less than the actual price.
Hence, this is the required solution.
This is what i got when i worked it out
