<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>
1.) y+6=3(x+2) C
y+6=3x+6
y=3x+6-6
y=3x+6
2.) y=1/2(x+8)-2 B
y=1/2x+4-2
y=1/2x+2
3.) y+1=1(x-3) E
y+1=x-3
y=x-3-1
y=x-4
4.) -4x+y=-2 A
y=-4x-2
5.) 2x-4y=-4 F
-4y=-2x-4
y=-2/-4x-4/-4
y=2/4x+4/4
y=1/2x+1
6.) 2x+4y=8 D
4y=-2x+8
y=-2/4x+8/4
y=-1/2x+4/2
y=-1/2x+2
Answer: Each piece would be 1/4 or 25%
Step-by-step explanation:
Step-by-step explanation:
let us give all the quantities in the problem variable names.
x= amount in utility stock
y = amount in electronics stock
c = amount in bond
“The total amount of $200,000 need not be fully invested at any one time.”
becomes
x + y + c ≤ 200, 000,
Also
“The amount invested in the stocks cannot be more than half the total amount invested”
a + b ≤1/2 (total amount invested),
=1/2(x + y + c).
(x+y-c)/2≤0
“The amount invested in the utility stock cannot exceed $40,000”
a ≤ 40, 000
“The amount invested in the bond must be at least $70,000”
c ≥ 70, 000
Putting this all together, our linear optimization problem is:
Maximize z = 1.09x + 1.04y + 1.05c
subject to
x+ y+ c ≤ 200, 000
x/2 +y/2 -c/2 ≤ 0
≤ 40, 000,
c ≥ 70, 000
a ≥ 0, b ≥ 0, c ≥ 0.
