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Luda [366]
3 years ago
13

Use a proof by contradiction to prove that the sum of two odd integers is even CM

Mathematics
1 answer:
9966 [12]3 years ago
8 0

Answer:

Sum of two odd integers is always even.

Step-by-step explanation:

Let m and n be two odd integers.

Since m and n are odd they can be written in the form m =2r + 1 and n = 2s + 1, where r and s are integers.

Let us suppose that their sum is not even.

m + n = (2r+1) + (2s + 1)

          = 2r + 2s + 2

          = 2(r+s+1)

          = 2z

Thus, the sum of m and n can be written in the form 2z where z is an integer. But this is a contradiction to the fact that their sum is even.

Hence, our assumption was wrong and the sum of two odd integers is always even.

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amm1812

Answer:

Step-by-step explanation:

(1x/4 + 1/7) + (3x/8 - 1/3) = x/4 + 3x/8 + 1/7 - 1/3

                                      = x*2/4*2 + 3x/8 + 1*3/7*3 - 1*7/3*7

                                      = 2x/8 + 3x/8 + 3/21 - 7/21

                                      = (2x+3x)/8 + (3-7) /21

                                      = 5x/8 + (-4)/21

                                       = 5x/8 - 4/21

8 0
3 years ago
I need a quick answer , someone please help
devlian [24]
Measurement a is 45
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4 0
3 years ago
Ten slips of paper labeled from 1 to 5 are placed in a hat. The first slip of paper is not replaced before selecting the second
vagabundo [1.1K]

Answer:

1/10

Step-by-step explanation:

There are two numbers less than 3 (1 and 2), and one number greater than 4 (5).

When the first slip is drawn, there are five slips in the hat, so the probability of a number less than 3 is 2/5.

When the second slip is drawn, there are four slips in the hat, so the probability of a number greater than 4 is 1/4.

Therefore, the total probability is (2/5) (1/4) = 1/10

6 0
3 years ago
We are throwing darts on a disk-shaped board of radius 5. We assume that the proposition of the dart is a uniformly chosen point
Vlad1618 [11]

Answer:

the probability that we hit the bullseye at least 100 times is 0.0113

Step-by-step explanation:

Given the data in the question;

Binomial distribution

We find the probability of hitting the dart on the disk

⇒ Area of small disk / Area of bigger disk

⇒ πR₁² / πR₂²

given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1

so we substitute

⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04

Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.

so

X ~ Bin( 2000, 0.04 )

n = 2000

p = 0.04

np = 2000 × 0.04 = 80

Using central limit theorem;

X ~ N( np, np( 1 - p ) )

we substitute

X ~ N( 80, 80( 1 - 0.04 ) )

X ~ N( 80, 80( 0.96 ) )

X ~ N( 80, 76.8 )

So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;

we covert to standard normal variable

⇒ P( X ≥  \frac{100-80}{\sqrt{76.8} } )

⇒ P( X ≥ 2.28217 )

From standard normal distribution table

P( X ≥ 2.28217 ) = 0.0113

Therefore, the probability that we hit the bullseye at least 100 times is 0.0113

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Hshhshs ssu sid si sisis s   sis s  s s is is sis si sis s s isis s i si
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