Two stores have movies to rent
2 answers:
Answer:
<h2>7</h2>Step-by-step explanation:
We know that the first store charges $12.50 per month, which is a initial condition, and charges additionally $1.50 per movie, which is variable, this represents the ratio of change, so this can be expressed as
Where represents movies.
Now, the second store doesn't charge and membership fee, just it charges a cost per movie which is $3.50.
Then, to solve the minimum number of movies needed to Plan A be the best choice, we just have to solve the following inequality
Which expresses the case where Plan A is a better choice, solving for , we have
Which means that the minimum number of movies is 7, which is the next whole number after 6.
You might be interested in
Answer:
(a) The solution is x=47.
(b) The solution is x=223755.
(c) The solution is x=33703314.
(d) The solution is x=984414.
Step-by-step explanation:
(a) Step 1 is to solve
q e
2 4 265 250 Calculation I
3 3 374 335 Calculation II
Now Solving for calculation I:
x≡≡
Solve (265)x=250(mod 433) for x0,x1,x2,x3.
x0:(26523)x0=25023(mod 433)⟹(432)x0=432⟹x0=1
x1:(26523)x1=(250×265−x0)22(mod 433)=(250×265−1)22(mod433)=(250×250)22(mod 433)⟹(432)x1=432⟹x1=1
x2:(26523)x2=(250×265−x0−2x1)21(mod 433)=(250×265−3)22(mod 433)=(250×195)21(mod 433)⟹(432)x2=432⟹x2=1
x3:(26523)x3=(250×265−x0−2x1−4x2)20(mod 433)=(250×265−7)20(mod 433)=(250×168)20(mod 433)⟹(432)x3=432⟹x3=1
Thus, our first result is:
x≡x0+2x1+4x2+8x3(mod24)≡1+2+4+8(mod24)≡15(mod24)
Now for Calculation II:
x≡≡
Solve (374)x=335(mod 433) for x0,x1,x2.
x0:(37432)x0=33532(mod 433)⟹(234)x0=198⟹x0=2. Note: you only needed to test x0=0,1,2, so it is clear which one x0 is.
x1:(37432)x1=(335×374−x0)31(mod 433)=(335×374−2)31(mod 433)=(335×51)31(mod 433)=1(mod 433)⟹(234)x1=1(mod 433)⟹x1=0
x2:(37432)x2=(335×374−x0−3x1)30(mod 433)=(335×374−2)30(mod 433)=(335×51)30(mod 433)=198(mod 433)⟹(234)x2=198(mod 433)⟹x2=2. Note: you only needed to test x2=0,1,2, so it is clear which one x2 is.
Thus, our second result is:
x≡x0+3x1+9x2(mod 33)≡2+0+9×2(mod 33)≡20(mod 33)
Step 2 is to solve
x ≡15 (mod 24 ),
x ≡20 (mod 33 ).
The solution is x=47.
(b) Step 1 is to solve
q e
2 10 4168 38277 523
3 6 674719 322735 681
is calculated using same steps as in part(a).
Step 2 is to solve
x ≡ 523 (mod 210 ),
x ≡ 681 (mod 36 ).
The solution is x=223755 .
(c) Step 1 is to solve
q e
2 1 41022298 1 0
29 5 4 11844727 13192165
In order to solve the discrete logarithm problem modulo 295 , it is best to solve it step by step. Note that 429 = 18794375 is an element of order 29 in F∗p . To avoid notational confusion, we use the letter u for the exponents.
¢294
First solve 18794375u0 = 11844727
= 987085.
The solution is u0 = 7.
The value of u so far is u = 7.
¢293
Solve 18794375u1 = 11844727·4−7
= 8303208.
The solution is u1 = 8.
The value of u so far is u = 239 = 7 + 8 · 29.
¢292
Solve 18794375u2 = 11844727 · 4−239
= 30789520.
The solution is
u2 = 26. The value of u so far is u = 22105 = 7 + 8 · 29 + 26 · 292 .
¢291
Solve 18794375u3 = 11844727 · 4−22105
= 585477.
The solution is
u3 = 18. The value of u so far is u = 461107 = 7 + 8 · 29 + 26 · 292 + 18 · 293 .
¢290
Solve 18794375u4 = 11844727 · 4−461107
= 585477.
The solution is
u4 = 18. The final value of u is u = 13192165 = 7 + 8 · 29 + 26 · 292 + 18 · 293 + 18 · 294 , which is the number you see in the last column of the table.
Step 2 is to solve
x ≡ 13192165 (mod 295 ).
x ≡ 0 (mod 2),
The solution is x=33703314 .
(d) Step 1 is to solve
q e
2 1 1291798 1 0
709 1 679773 566657 322
911 1 329472 898549 534
To solve the DLP’s modulo 709 or 911, they can be easily solved by an exhaustive search on a computer, and a collision algorithm is even faster.
Step 2 is to solve
x ≡ 0 (mod 2),
x ≡ 322 (mod 709),
x ≡ 534 (mod 911).
The solution is x=984414