Which expression is equivalent to StartFraction 3 x Over x + 1 EndFractiondivided by x + 1?

If we compare the p value to a predetermined significance level we conclude that we cannot reject the null hypothesis, indicating insufficient evidence to assert that the true average is different from 6.6 pounds.

The best conclusion is:

a. Failing to reject the null hypothesis.

La proporción de ambas velas es 20/60, que simplificada es 1/3, por lo que debes tomar la cantidad de onzas que tiene la vela y dividirla por 1/3. Dicho de otra manera: 9 / 1/3 => 9 * 3 = 18

Answer:

1. x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

By extracting the square root on both sides

x=±4

2. t^2=81

Again, take the square root of each side

t=±9

3. r^2-100=0

r=±10

4. x²-144=0

We rewrite as x²=144

Applying square roots

x=±12

5. 2s²=50

s=±5 ..

Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

The determined Z value is 2.53

In this case, the critical value Z is 1.96 at the 0.05 significance level

Step(iii):-

Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

We accept the alternative hypothesis

Conclusion:-

We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds

Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R displays antisymmetry, as whenever (a,b)∈R, it follows that a=b.

R lacks reflexivity since (1,1) ∉ R even though 1 ∈ A.

R is transitive; therefore, if (a,b)∈R and (b, c) ∈ R, then a=b=c and (a,c)=(a,a)∈R.

R fails to be a partial ordering due to its lack of reflexivity.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric because if (a,b)∈R and (b, a) ∈ R, then a must equal b (e.g., (2,0) ∈ R and (0,2) ∉ R; likewise, (2,3) ∈ R and (3,2) ∉ R).

R is reflexive since each (a,a) resides in R for all elements a ∈ A.

R is transitive; if (a,b)∈R and (b,c)∈R, it implies (a,c) exists in R or identical to (a,b) in R.

R qualifies as a partial ordering due to its reflexivity, antisymmetry, and transitivity.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive as (a,a)∈R is true for every a ∈ A.

R is antisymmetric; if (a,b)∈R holds and if also (b,a)∈R, then a invariably equals b (e.g., (1,2)∈R while (2,1) ∉ R; similarly for (3,1) and (1,3)).  

R fails transitivity because (3,1) ∈ R and (1,2) ∈ R, but (3,2) ∉ R.

R is not a partial ordering due to transitivity not being satisfied.

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R exhibits reflexivity since (a,a)∈R for each element a ∈ A.

R displays antisymmetry, as if (a,b)∈R and (b,a)∈R then a must equal b (e.g., (1,2)∈R and (2,1)∉R; similarly validated for others).

R is not transitive because (1,2)∈R and (2,0)∈R, but (1,0)∉R.

R is not a partial ordering due to transitivity issues.

e):  R = { ( 0, 0 ), ( 0, 1 ), ( 0, 2 ), ( 0, 3 ), ( 1, 0 ), ( 1, 1 ), ( 1, 2 ), ( 1, 3 ), ( 2, 0 ), ( 2, 2 ), ( 3, 3 ) }

R proves to be reflexive, given that (a,a)∈R for all a∈A.

R is not antisymmetric since both (1,0)∈R and (0,1)∈R hold while 0 is distinct from 1.

R lacks transitivity, as (2,0)∈R and (0,3)∈R, while (2,3)∉R.

R cannot be classified as a partial ordering as it fails in both antisymmetry and transitivity.