Mr. Smith runs a sprinkler 150 m in 14.98 seconds Steve runs his sprinkler 100 m in 10.23 seconds who’s sprinkler is faster

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

5 0

2 years ago

What are the quotient and remainder when 777 is divided by 21?

DIA [1.3K]

777 divided by 21 = 34 with a remainder of 3

5 0

2 years ago

Subtract the following polynomials, then place the answer in the proper location on the grid.

Grace [21]

(3m - 4n - 7) - (8m + n - 6)

3m - 8m = - 5m
-4n - n = -5n
-7 - (-6) = -7 + 6 = -1

-5m - 5n - 1 is your answer

hope this helps

8 0

3 years ago

In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until

Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this problem:

There are 12 bulbs, hence N = 12.

3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.

The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182$

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0

2 years ago

WILL GIVE BRAINLIEST IF CORRECT ANSWER!!!

Rom4ik [11]

Answer : 2√3

Given :-

A equilateral triangle with side length 4.

To Find :-

The value of x in the given figure.

As we know that in a equilateral triangle , perpendicular bisector , angle bisector and median coincide with each other .

So the perpendicular drawn in the figure will bisect the given side .
Therefore the value of each half will be 4/2 = 2 .

Now we may use Pythagoras theorem as ,

→ AB² = BC² + AC²

→ 4² = 2² + x²

→ 16 = 4 + x²

→ x² = 16-4

→ x² = 12

→ x =√12 = √{ 3 * 2²}

→ x = 2√3

Hence the required answer is 2√3 .

I hope this helps.

3 0

2 years ago