The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen

Answer:

• Yes; refer to the question's clarification and the detailed answer below for more information.

Explanation:

Sets are in bijection if there is a bijective function connecting them.

This implies that the question can be interpreted as: can sets A and B be related via a bijective function?

A bijective function must be both injective and surjective, which means it is reversible.

Injective indicates that each element in the domain corresponds uniquely to an element in the codomain.

Surjective denotes that every element in the codomain relates to exactly one element in the domain.

This ensures there are no unmatched elements.

If we consider set A as the function's domain and set B as the codomain.

The core question then becomes: can the set {17, 34, 51, 68, 85,...} be in bijection with the set {11, 22, 33, 44, 55,...}?

Notice that the two sets share no common elements, as 17 and 34 are coprime.

A bijective function can only be established between sets A and B or B and A if both sets possess the same cardinality (number of elements).

Indeed, the cardinality of sets A and B is identical: both sets are of equal size, affirming that a bijective relationship exists between them.

You can acquire 42 cookies through 12 different combinations. The first method involves purchasing 2 packs of 21 (21x2 = 42). The second consists of acquiring 1 pack of 21 alongside 3 packs of 7 (21 + 3x7 = 42). The third way is to buy 1 pack of 21 and 21 individual cookies (21 + 21 = 42). The fourth option combines 1 pack of 21, 1 pack of 7, and 14 single cookies (21 + 7 + 14 = 42). The fifth strategy includes 1 pack of 21, 2 packs of 7, and 7 individual cookies (21 + 14 + 7 = 42). The sixth way is to opt for 6 packs of 7 (7x6 = 42). The seventh option is to purchase 5 packs of 7 along with 7 individual cookies (7x5 + 7 = 42). For the eighth method, you can buy 4 packs of 7 and 14 single cookies (7x4 + 14 = 42). The ninth way is to get 3 packs of 7 with 21 single cookies (7x3 + 21 = 42). The tenth consists of acquiring 2 packs of 7 plus 28 individual cookies (7x2 + 28 = 42). The eleventh strategy involves 1 pack of 7 and 35 single cookies (7 + 35 = 42). Lastly, the twelfth method is simply buying 42 individual cookies (42 = 42).

The equation of the perpendicular line can be identified by determining its slope and applying the given point within the standard formula.

Standard equation: y-y1 = m(x-x1)

m*m'=-1
where m' indicates the slope of the perpendicular line
m denotes the slope of the original line

m = -coefficient of x/coefficient of y = -4/-3 = 4/3
m' = -3/4

Substituting the point (3, -2):
y+2 = -3/4*(x-3)
4y+8 = -3x+9

Thus, the equation of the perpendicular line is: 3x+4y-1=0

To work with the repeating decimals, note that since there are two digits repeating, you will be using a 1 followed by two zeros, which is 100. This factor will be important for subsequent calculations.

Let’s denote the repeating decimal as x.

Thus, you can write

x = 0.212121...

Now, if you multiply this equation by 100, you can write it on top of the previous equation.
This results in:

100x = 21.212121...
x = 0.212121...

Next, set up a subtraction by taking the second equation from the first:

100x = 21.212121...
x = 0.212121...
- ----------------------------
99x = 21

Solving for x involves dividing both sides by 3:

33x = 7

Then, divide both sides by 33:

x = 7/33

Thus, we conclude that the repeating decimal 0.212121... equates to the fraction 7/33.

Answer: the fraction is 7/33

Answer:

We require a total of 144 birthday candles for the 12 parties.

Step-by-step explanation:

As a caterer focused on children’s birthday celebrations, you have 12 events to cater to in the upcoming week, each requiring 2 cakes. Each cake will be adorned with 6 candles.

Thus, for all 12 birthday gatherings, you need a total of =

= 24 cakes.

Further, since every cake holds 6 candles, the overall number of candles for the 24 cakes amounts to =

= 144 candles.

Therefore, you will need 144 birthday candles to accommodate the 12 celebrations.