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Andrej [43]
3 years ago
15

A formal proof uses all of the following components except what?

Mathematics
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

A.

Step-by-step explanation:

You might be interested in
Bethany wrote the equation X+ (x+2)+(+4)= 91 when she was told that the sum of three consecutive odd integers had a
azamat

Answer:

Option A) Bethany is correct because consecutive odd integers will each have a difference of two

Step-by-step explanation:

The sum of 3 consecutive odd integers is 91. Let the first odd integer is x. The next odd integer will be obtained by adding 2 in x i.e. (x + 2). The third odd integer will be obtained by adding 2 in the second odd integer i.e. (x + 2) + 2 = x + 4

So, the 3 odd integers will be:

x  , (x+2) and (x+4)

Their sum is given to be 91. So we can write:

x + (x+2) + (x+4) = 91

Hence, we can conclude that: Bethany is correct because consecutive odd integers will each have a difference of two.

Other options are not correct because consecutive odd integers always increase by 2. For example, the next odd integer after 1 is 3, which is obtained by adding two, similarly the next odd will be 5 and so on.

5 0
4 years ago
Read 2 more answers
Here's another question, same one (kind of) <br> 5x - 8 = 3x + 4
cluponka [151]

Answer:

x=6

Step-by-step explanation:

5x - 8 = 3x + 4

    +8         +8

5x=3x+12

-3x  -3x

2x=+12

/2    /2

x=6

6 0
3 years ago
] Consider a randomly shuffled deck of ten cards labeled {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three to be
ollegr [7]

Answer:

B should be worth more points.

Step-by-step explanation:

Given -  Consider a randomly shuffled deck of ten cards labeled

              {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, of which game play requires three

               to be drawn.

To find - Which one should be worth more points:

              A getting a pair of matching cards within the three,

              or B getting a set of three cards that can be arranged such

              that the sum of two of them is the value of the third.

Proof -

Given that, three cards has to be drawn.

So, total number of ways the card be drawn = ¹⁰C₃

                                                                      = \frac{10!}{3! (10-3)!}

                                                                      = \frac{10!}{3! 7!}

                                                                      = \frac{10.9.8.7!}{3! 7!}

                                                                      = \frac{10.9.8}{3.2.1}

                                                                      = 120

⇒¹⁰C₃ = 120

Now,

Given that, A getting a pair of matching cards within the three

Sample space = { (1,1,2), (1,1,3).(1,1,4),(1,1,5);

                            (2,2,1),(2,2,3),(2,2,4),(2,2,5);

                             (3,3,1),(3,3,2),(3,3,4),(3,3,5);

                              (4,4,1),(4,4,2),(4,4,3),(4,4,5);

                              (5,5,1),(5,5,2),(5,5,3),(5,5,4) }

⇒n(A) = 20

Also,

B getting a set of three cards that can be arranged such that the sum of two of them is the value of the third

Sample space = { (1,1,2),(1,2,3),(1,3,4),(1,4,5),(2,2,4),(2,3,5) }

⇒n(B) = 6

Now,

P(A) = \frac{20}{120} = \frac{1}{6} = 0.167

P(B) = \frac{6}{120} = \frac{1}{20} = 0.05

We can see that, P(B) < P(A)

⇒B should be worth more points.

4 0
3 years ago
When is the mean most useful in describing a set of data?
Anettt [7]

Answer:

When the range is wide

Step-by-step explanation:

8 0
3 years ago
Can someone please help me ? Thank youu !
GalinKa [24]

Answer:

B

Step-by-step explanation:

5 0
3 years ago
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