Segment AB is congruent to segment AB.t h s statement shows the _____

For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)

vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

The relation being reflexive (∀a∈R, a∼a)
The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0

3 years ago

Geometry please help

JulsSmile [24]

C; 50 is the answer, since two of the sides are congruent...100/2 is 50

3 0

3 years ago

Nathan usually drinks 38 ounces of water per day. He read that he should drink 60 ounces of water per day. If he starts drinking

lesya692 [45]

Answer:

$58\%$

Step-by-step explanation:

First, we need to find how much extra water Nathan drinks.

$60-38=22$ extra ounces.

To find the percent increase:

$\frac{22}{38} =58\%$ (to the nearest percent)

Hope this helps :)

7 0

3 years ago

I am having trouble figuring out this problem. I believe the answer is (0,7) but I am not sure.

AleksandrR [38]

It is (0,7).

7(cos(pi/2)+isin(pi/2))

7i which is (0, 7)

7 0

3 years ago

Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disorde

zmey [24]

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

$p_v =P(t_{(19)}>1.638)=0.0589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation

$n=20$ sample size

$\mu_o =3.58$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 3.58 :

Null hypothesis: $\mu \leq 3.58$

Alternative hypothesis: $\mu > 3.58$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since is a right tailed test the p value would be:

$p_v =P(t_{(19)}>1.638)=0.0589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

7 0

3 years ago

Segment AB is congruent to segment AB.t h s statement shows the ______ property.