Answer:
The price of the cell phone without the coupon= $500
Step-by-step explanation:
Step 1: Express discounted amount
The discounted amount can be expressed as a function of the original cost of the phone as follows;
D=r×A
where;
D=discounted amount
r=coupon rate
A=original price of the cell phone before the coupon
In our case;
r=45%=45/100=0.45
A=a
replacing;
Discounted amount=(0.45×a)=0.45 a
Step 2: Amount she pays up
Amount she pays=Original cost of cell phone-discounted amount
where;
Amount she pays= $275
original cost of cell phone=a
discounted amount=0.45 a
replacing;
$275=a-0.45 a
0.55 a=275
a=275/0.55
a=500
The price of the cell phone without the coupon= $500
The thousands place (1000) is the fourth number to the left of the decimal, so round that one either up or down based on the number just one spot closer to the decimal, in this case, the hundreds place (100).
so we have 32,420
we can ignore the numbers that are going to become zeros as a result of the rounding . .
32,400
let's round 32,XXX either up or down based on the value in the hundreds place. if that value is 5 or greater, than we round up. if it is less than 5, than we round down.
4 is in the hundreds place, so we round down
32,000 is the answer
P(defective) = 3/12 = 1/4
P(good) = 1 - 1/4 = 3/4
The probability that at least 2 units are good is given by:
P(2 good) + P(3 good) + P(4 good) = 0.211 + 0.422 + 0.316 = 0.949.
Answer:
Step-by-step explanation:
Given:
∠DCE ≅ ∠DEC
∠B ≅ ∠F
DF ≅ BD
To prove:
ΔABC ≅ ΔGFE
Solution:
Statements Reasons
1). ∠DCE ≅ ∠DEC 1). Given
2). ∠ACB ≅ ∠GEF 2). Vertically opposite angles to the
congruent angles.
3). ∠B ≅ ∠F 3). Given
4). DB ≅ DF 4). Given
5). DC + CB ≅ DE + EF 5). Segment addition postulate
6). DC ≅ DE 6). Property of isosceles triangle
7). CB ≅ EF 7). Transitive property
8). ΔABC ≅ ΔGFE 8). ASA property of congruence
