Answer:
![\huge\boxed{r_{x-axis}(x,\ y)\to(x,\ -y)}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7Br_%7Bx-axis%7D%28x%2C%5C%20y%29%5Cto%28x%2C%5C%20-y%29%7D)
Step-by-step explanation:
Read the coordinates of the points:
A(-1, 0) → A'(-1, 0)
B(0, 2) → B'(0, -2)
C(3, 2) → C'(3, -2)
D(4, 0) → D'(4, 0)
(x, y) → (x, -y)
Check:
A → A': (-1, 0) → (-1, -0) = (-1, 0) CORRECT
B → B': (0, 2) → (0, -2) CORRECT
C → C': (3, 2) → (3, -2) CORRECT
D → D': (4, 0) → (4, -0) = (4, 0) CORRECT
Answer:
The linear equation used to find the cost to rent the shoes is ![24.75+x=32](https://tex.z-dn.net/?f=24.75%2Bx%3D32)
Cost to rent the shoes is $7.25.
Step-by-step explanation:
Given:
Cost of per game of bowling = $4.95
Number of games played = 5
Total money paid = $32
We need to find the cost to rent the shoes.
Solution:
Let the cost to rent the shoes be 'x'.
So we can say that;
Total money paid is equal to sum of Cost of per game of bowling multiplied by Number of games played and cost to rent the shoes.
framing in equation form we get;
![5\times4.95+x=32\\\\24.75+x=32](https://tex.z-dn.net/?f=5%5Ctimes4.95%2Bx%3D32%5C%5C%5C%5C24.75%2Bx%3D32)
Hence The linear equation used to find the cost to rent the shoes is ![24.75+x=32](https://tex.z-dn.net/?f=24.75%2Bx%3D32)
On solving the above equation we will find the value of 'x'.
Now we will subtract both side by 24.75 we get;
![24.75+x=32-24.75\\\\x= \$7.25](https://tex.z-dn.net/?f=24.75%2Bx%3D32-24.75%5C%5C%5C%5Cx%3D%20%5C%247.25)
Hence Cost to rent the shoes is $7.25.
Answer:
HiStep-by-step explanation:
J
6x6 = 36
1/2(6)(6) = 18
1/2(6)(6) = 18
36+18+18 = 72
answer is D. 72 in^2
Answer: 0.0325
Step-by-step explanation:
Binomial probability distribution formula for x successes in n trials:-
, where n is the number of trials , p is the probability of success.
Given : The probability that taxpayers who filed their tax return are electronically self-prepared their taxes : ![p= 0.319](https://tex.z-dn.net/?f=p%3D%200.319)
If three tax returns submitted electronically are randomly selected, then the probability that all three were self-prepared is given by :-
![P(X=3)=^3C_3\ (0.319)^3\ (1-0.319)^{3-3}\\\\=(1) (0.319)^3(1)\\=0.032461759\approx0.0325](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5E3C_3%5C%20%280.319%29%5E3%5C%20%281-0.319%29%5E%7B3-3%7D%5C%5C%5C%5C%3D%281%29%20%280.319%29%5E3%281%29%5C%5C%3D0.032461759%5Capprox0.0325)
Hence, the probability that all three were self-prepared =0.0325