Theresa rolled a number cube, number 1 through 6, two times

Carla can choose two of her three pairs of sneakers to take to a track meet. If the pairs of sneakers are called A, B, and C, wh

lawyer [12517]

The formula for determining the number of ways to select two pairs of sneakers from three pairs labeled A, B, and C is represented by 3C2 (3 combinations of 2), which calculates as 3! / (2!(3 - 2)!) = 3! / (2! x 1!) = (3 x 2) / (2 x 1) = 3. Therefore, the sample space is S = {AB, AC, BC}.

5 0

1 month ago

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In Andrew’s Furniture Shop, he builds bookshelves and tables. Each type of furniture takes him about the same time to make. He f

babunello [11817]

A. The order in which you define the variables isn't fixed. For the sake of this discussion, let's define them like this:

x = Number of bookshelves
y = Number of tables

B. Due to the total number of items to produce, we have the following inequality based on those variables.

x + y > 25

Additionally, you can create a second inequality concerning your budget for materials.

20x + 45y < 675

Moreover, you should also add that both values must not be negative, since you can't produce negative tables.

C. By analyzing the constraints and solving the system, you will find that the feasible region contains 4 vertices.

(0,0)
(18, 7)
(0, 15)
(33.75, 0) or (33, 0) if you prefer to round it.

8 0

1 month ago

In the figure, CD = EF and AB = CE. Complete the statements to prove that AB = DF.

Svet_ta [12734]

Answer:

1. Addition Property of Equality

2. Segment addition

3. Substitution Property of Equality.

4. Transitive Property of Equality.

Step-by-step explanation:

Given: CD = EF and AB = CE

To Show: AB = DF

Following the steps outlined:

1. CD + DE = EF + DE by the (addition) Property of Equality.

This shows that the same value has been added to both sides of the equation.

2. CE = CD + DE and DF = EF + DE through (segment addition).

This indicates that CE and DF are segments, and the length of any segment equals the total of its parts.

3. CE = DF according to the (Addition, subtraction, substitution, transitive) Property of Equality. Recognizing that CE = CD + DE and using the fact that CD = EF allows us to replace CD with EF.

So, CE = EF + ED = FD (via substitution) Property of Equality.

4. Since AB = CE and CE = DF, we can conclude AB = DF by the (transitive) Property of Equality.

This follows the rule that if x = y and y = z, then x = z.

3 0

2 months ago

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Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t

AnnZ [12381]

Answer: a) $\bold{\dfrac{3}{16}}$ b) $\bold{\dfrac{1}{36}}$

Step-by-step explanation:

a) To achieve an even total, there are 3 possible combinations:

1) Even, Even, Even, Even $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

2) Even, Even, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

3) Odd, Odd, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

Order is irrelevant

Summing these yields your final result: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die shows a 2 and another a 3, while the remaining two can show any digits, there’s only one way to get a 2, one way for a 3, and six potential numbers for each of the other two dice.

$\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}$

3 0

2 months ago

Find three different surfaces that contain the curve r(t) = t^2 i + lnt j + (1/t)k

Svet_ta [12734]

Let’s consider the curve: r(t) = t²i +(int)j + 1/t k X = t², y = int,z = 1/t Utilizing x = t² and z = 1/t X = (1/z)² Xz² = 1 Now using y = int and z= 1/t Y = in│1/z│ By using x = t² and y = int Y = int = in(√x) Thus, the resulting surfaces are, Xz² = 1 Y = in│1/z│ Y= in(√x)

7 0

2 months ago