Fairly straightforward question. She works 30 days and is late 80% or 0.8
We can model our equation
Num days late = late percentage * num working days
Basically fine portion of working days that she was late.
Num days late = 0.8 * 30=24
So Deborah was late 24 out of 30 working days. Would appreciate a brainliest
we know that
The simple interest formula is equal to
where
P is the Principal amount of money to be invested
I is the amount of money in interest
r is the rate of interest
t is Number of Time Periods
in this problem we have
substitute in the formula above and solve for P
![P=14.65/[(0.025)(2)]](https://tex.z-dn.net/?f=P%3D14.65%2F%5B%280.025%29%282%29%5D)
therefore
<u>the answer is</u>
Sorry if this is wrong and it doesn't hep but I got 32
Answer:
x=-1, y=-3. (-1, -3).
Step-by-step explanation:
x+y=-4
3x-6y=15
---------------
x=-4-y
3(-4-y)-6y=15
-12-3y-6y=15
-12-9y=15
9y=-12-15
9y=-27
y=-27/9
y=-3
x+(-3)=-4
x-3=-4
x=-4+3
x=-1
Step-by-step explanation:
a). A = {x ∈ R I 5x-8 < 7}
5x - 8 < 7 <=> 5x < 8+7 <=> 5x < 15 =>
x < 3 => A = (-∞ ; 3)
A ∩ N = {0 ; 1 ; 2}
A - N* = (-∞ ; 3) - {1 ; 2}
b). A = { x ∈ R I 7x+2 ≤ 9}
7x+2 ≤ 9 <=> 7x ≤ 7 => x ≤ 1 => x ∈ (-∞ ; 1]
A ∩ N = {0 ; 1}
A-N* = (-∞ ; 1)
c). A = { x ∈ R I I 2x-1 I < 5}
I 2x-1 I < 5 <=> -5 ≤ 2x-1 ≤ 5 <=>
-4 ≤ 2x ≤ 6 <=> -2 ≤ x ≤ 3 => x ∈ [-2 ; 3]
A ∩ N = {0 ; 1 ; 2 ; 3}
A - N* = [-2 ; 3) - {1 ; 2}
d). A = {x ∈ R I I 6-3x I ≤ 9}
I 6-3x I ≤ 9 <=> -9 ≤ 6-3x ≤ 9 <=>
-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3 =>
-3 ≤ x ≤ 5 => x ∈ [-3 ; 5]
A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}
A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}
