# Which numbers are solutions to the inequality x greater-than 14 and one-half? Check all that apply.

###### Question:

fractions larger than 14 and one-half

decimals larger than 14 and one-half

whole numbers larger than 14 and one-half

the number 14 and one-half

fractions smaller than 14 and one-half

decimals smaller than 14 and one-half

whole numbers smaller than 14 and one-half

## Answers

A, B, And C

Step-by-step explanation:

Hop this helps! <3

fractions larger than 14 and one-half

decimals larger than 14 and one-half

whole numbers larger than 14 and one-half

(A,B,C (: hope this helps

Step-by-step explanation:

1/2b would be a number half as big as b.

(1,3,6,1,3,3)

BrO3-(aq) + 3Sn2+(aq) + 6H+(aq) → Br-(aq) + 3Sn4+(aq) + 3H2O(l)

Explanation:

BrO3-(aq) + Sn2+(aq) --> Br-(aq)+Sn4+(aq)

Separating into half equations, we have;

Sn2+ → Sn4+ + 2e- (Oxidation)

BrO3- + 6e- → Br- (Reduction)

Upon balancing the atoms and charges we have;

Sn2+ → Sn4+ + 2e- (Oxidation)

BrO3- + 6e- + 6H+ → Br- + 3H2O (Reduction)

Make sure electron gain is equal to electron lost, we have;

3Sn2+ → 3Sn4+ + 6e- (Oxidation)

BrO3- + 6e- + 6H+ → Br- + 3H2O (Reduction)

Add both half equations together;

BrO3- + 6e- + 6H+ + 3Sn2+ → Br- + 3H2O + 3Sn4+ + 6e-

Upon simplification;

BrO3-(aq) + 3Sn2+(aq) + 6H+(aq) → Br-(aq) + 3Sn4+(aq) + 3H2O(l)

What are the coefficients of the six species in the balanced equation above?

(1,3,6,1,3,3)

3 /4 x+ 1 /2 one-half more than three-fourths of a number

3/ 4 − 1 /2 x three-fourths minus one-half of a number

3/ 4 −(x+ 1/ 2 ) three-fourths minus the sum of a number and one-half

a) 0.425

b) 0.65

c) 0.575

Step-by-step explanation:

The ratio of even balls to odd balls is 3:7. That means if there are 3 even balls, then there are 7 odd balls. That makes a total of 10 balls, of which 30% are even and 70% are odd.

a) There are 240 balls, so originally, 72 are even and 168 are odd. 30 of the odd balls are renumbered by multiply by 4. An odd number times an even number is an even number, so 30 of the odd balls become even balls. So there are now 102 even balls and 138 odd balls. Therefore:

P(even) = 102 / 240

P(even) = 0.425

b) Originally, 70% of the balls are odd. Half of those are renumbered by multiplying by 6, so they become even numbers.

P(odd) = 0.5 (70%) = 35%

P(even) = 30% + 35% = 65%

A third of the even balls are renumbered by multiplying by 5, so they remain even numbers.

Therefore, P(even) = 0.65.

c) Originally, 30% of the balls are even. Half of those are renumbered by adding 3. An even number plus an odd number is an odd number, so:

P(even) = 0.5 (30%) = 15%

P(odd) = 70% + 15% = 85%

Next, half the odd balls are renumbered by adding 5. An odd number plus an odd number is an even number, so:

P(odd) = 0.5 (85%) = 42.5%

P(even) = 15% + 42.5% = 57.5%

Therefore, P(even) = 0.575.

double half (double number) //to pass by reference is not important here

{

double half; //number again should not be declared here

half = number / 2;

return half;

}

int main()

{

cout<<half(50);

}

Output :

25

Explanation:

As not mentioned in the question, any data type could have been used like int , float or double. Method half() takes a double variable as parameter which is passed as value as we don't need to make the change at the same address we need value of that argument only. A variable half of double type is declared which will hold the half value. Number is not required to declared again here as it is passed by argument. Then the number is divided by 2 and half is returned.

OH−(aq), and H+(aq)

Explanation:

Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;

MnO4^-(aq) + 8H^+(aq) +5e> Mn^2+(aq) + 4H2O(l)

The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.

For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.

Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.

Part A: (1, 1, 4, 1, 1, 1)

Part B: (2, 6, 4, 2, 3, 8)

Explanation:

Redox reactions can be balanced using the half-reaction method. It has the following steps:

We write both half-reactions (reduction and oxidation)We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.We add electrons to balance electrically the half-reactionWe multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.We add both half-reactions and take the numbers to the general equation.Acidic solution

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Basic solution

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

So we conclude that the answer is under (B).

Step-by-step explanation:

We know that the penny is a fair coin; that is, the probability of heads is one-half and the probability of tails is one-half.

So when we throw a coin we have an equal chance of getting either a head or a tail.

So we conclude that the answer is under (B).

B) regardless of the number of flips, half will be heads and half tails.