💡To determine the probability that Melinda can form a two-digit number between 20 and 30 inclusive by rolling two standard six-sided dice, we need to identify all possible two-digit numbers within this range and calculate the probability of rolling the required digits.First, list the two-digit numbers between 20 and 30 inclusive:- 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30Next, determine the possible outcomes for each number:- For numbers 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, and 30, the first die must show a 2 or 3, and the second die must show a digit from 0 to 9. However, since we are using standard six-sided dice, the second die can only show digits from 1 to 6.Therefore, the valid two-digit numbers that can be formed are:- 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36Now, calculate the probability:- There are 12 valid outcomes out of a total of 36 possible outcomes when rolling two dice.Thus, the probability is:frac{12}{36} = frac{1}{3}However, upon reviewing the initial problem statement, it specifies that Melinda will make a two-digit number between 20 and 30 inclusive. This means we should only consider numbers from 20 to 30, which are:- 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30But since the second die can only show digits from 1 to 6, the valid numbers are:- 21, 22, 23, 24, 25, 26This gives us 6 valid outcomes. Therefore, the probability is:frac{6}{36} = frac{1}{6}However, this contradicts the initial calculation. To resolve this, we need to consider that the first die must show a 2 to form numbers between 20 and 29, and a 3 to form 30. But since 30 requires the first die to show a 3 and the second die to show a 0, which is not possible with a standard die, 30 cannot be formed.Thus, the only valid numbers are 21, 22, 23, 24, 25, and 26, giving us 6 valid outcomes. Therefore, the correct probability is:frac{6}{36} = frac{1}{6}However, this still seems inconsistent with the initial thought process. To accurately determine the probability, we should consider that the first die must show a 2 to form numbers between 20 and 29. The second die can show any digit from 1 to 6, giving us 6 valid outcomes. Additionally, to form 30, the first die must show a 3 and the second die must show a 0, which is not possible. Therefore, the only valid numbers are 21, 22, 23, 24, 25, and 26, resulting in 6 valid outcomes.Thus, the probability is:frac{6}{36} = frac{1}{6}But this contradicts the initial conclusion of frac{11}{36}. To reconcile this, we need to consider that the initial approach might have included numbers beyond 30, which are not allowed. Therefore, the correct probability should be based on the valid numbers between 20 and 30 inclusive, which are 21, 22, 23, 24, 25, and 26, giving us 6 valid outcomes.Hence, the final probability is:frac{6}{36} = frac{1}{6}

💡Okay, so I need to find the sum of all integers that are greater than 2 and less than 15. Hmm, let me think about how to approach this. First, I should identify which numbers we're talking about. If the numbers are greater than 2, that means we start from 3. And if they're less than 15, we go up to 14. So, the numbers we're adding are 3, 4, 5, all the way up to 14.Now, adding all these numbers one by one seems tedious. There must be a smarter way to do this. I remember something about arithmetic series and how to find their sums without having to add each term individually. Let me recall the formula for the sum of an arithmetic series. I think it's something like S = n/2 * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term.Okay, so first, I need to figure out how many terms there are from 3 to 14. To do this, I can subtract the smallest number from the largest and then add 1 to include both endpoints. So, 14 minus 3 is 11, and adding 1 gives me 12 terms. That means n = 12.Next, I need to identify a1 and an. In this case, a1 is 3, and an is 14. Plugging these into the formula, I get S = 12/2 * (3 + 14). Simplifying that, 12 divided by 2 is 6, and 3 plus 14 is 17. So, now I have S = 6 * 17.Wait, what's 6 times 17? Let me calculate that. 6 times 10 is 60, and 6 times 7 is 42. Adding those together gives me 102. So, the sum of all integers greater than 2 and less than 15 is 102.But just to make sure I didn't make a mistake, let me verify this another way. Maybe I can add the numbers in pairs. If I pair the first and last terms, 3 and 14, their sum is 17. The next pair, 4 and 13, also adds up to 17. Continuing this pattern, 5 and 12 is 17, 6 and 11 is 17, 7 and 10 is 17, and 8 and 9 is 17. That's six pairs, each adding up to 17. So, 6 times 17 is indeed 102. That matches my earlier calculation, so I feel confident that 102 is the correct sum.I guess another way to think about it is using the formula for the sum of the first n natural numbers, which is S = n(n + 1)/2. But in this case, we're not starting from 1, so I need to adjust for that. The sum from 1 to 14 would be 14*15/2 = 105. The sum from 1 to 2 is 3. So, subtracting the sum from 1 to 2 from the sum from 1 to 14 gives me 105 - 3 = 102. That's another confirmation that the sum is 102.Alright, I think I've approached this problem from a few different angles and each time I arrive at the same answer. It seems pretty solid. I don't see any mistakes in my reasoning, so I'm ready to present the final answer.

💡First, I need to solve the equation (14x + 5 - 21x^2 = -2) to find the value of (x).I'll start by rearranging the equation to standard quadratic form:[-21x^2 + 14x + 5 = -2]Subtracting (-2) from both sides gives:[-21x^2 + 14x + 7 = 0]To simplify, I'll divide the entire equation by (-7):[3x^2 - 2x - 1 = 0]Now, I'll solve this quadratic equation using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, (a = 3), (b = -2), and (c = -1). Plugging these values into the formula:[x = frac{-(-2) pm sqrt{(-2)^2 - 4 cdot 3 cdot (-1)}}{2 cdot 3} = frac{2 pm sqrt{4 + 12}}{6} = frac{2 pm sqrt{16}}{6} = frac{2 pm 4}{6}]This gives two solutions:[x = frac{6}{6} = 1 quad text{and} quad x = frac{-2}{6} = -frac{1}{3}]Next, I'll substitute these values of (x) into the expression (6x^2 - 4x + 5) to find its value.For (x = 1):[6(1)^2 - 4(1) + 5 = 6 - 4 + 5 = 7]For (x = -frac{1}{3}):[6left(-frac{1}{3}right)^2 - 4left(-frac{1}{3}right) + 5 = 6left(frac{1}{9}right) + frac{4}{3} + 5 = frac{2}{3} + frac{4}{3} + 5 = 2 + 5 = 7]In both cases, the value of the expression is 7.

💡First, I need to determine the percentage of candidates selected from State A. Let's denote this percentage as ( x% ).Given that the number of candidates who appeared from each state is 8000, the number of candidates selected from State A can be expressed as:[frac{x}{100} times 8000]For State B, it's stated that 7% of the candidates got selected, and this number is 80 more than the number selected from State A. Therefore, the number of candidates selected from State B is:[frac{7}{100} times 8000 + 80]According to the problem, the number of candidates selected from State B is 80 more than those selected from State A. This leads to the equation:[frac{x}{100} times 8000 + 80 = frac{7}{100} times 8000]Simplifying the right side of the equation:[frac{7}{100} times 8000 = 560]Now, the equation becomes:[frac{x}{100} times 8000 + 80 = 560]Subtracting 80 from both sides:[frac{x}{100} times 8000 = 480]To solve for ( x ), divide both sides by 8000:[frac{x}{100} = frac{480}{8000}]Simplifying the fraction:[frac{x}{100} = 0.06]Multiplying both sides by 100 to find ( x ):[x = 6]Therefore, the percentage of candidates selected from State A is 6%.

💡First, I need to determine how many cards Norma had initially. She found 70.0 cards and now has a total of 158 cards.To find the initial number of cards, I can set up an equation where the initial number of cards plus the cards she found equals the total number of cards she has now.So, let ( x ) represent the initial number of cards. The equation would be:[x + 70.0 = 158]Next, I'll solve for ( x ) by subtracting 70.0 from both sides of the equation:[x = 158 - 70.0][x = 88]Therefore, Norma initially had 88 cards.

💡Alright, so I have this function ( f(x) = ax^2 + (2a + 1)x + b ), and there are three parts to the problem. Let me tackle each part one by one.**Part I: Finding the zeros when ( a = 1 ) and ( b = -4 ).**Okay, so if ( a = 1 ) and ( b = -4 ), the function becomes:( f(x) = x^2 + (2*1 + 1)x - 4 )Simplify that:( f(x) = x^2 + 3x - 4 )Now, to find the zeros, I need to solve ( x^2 + 3x - 4 = 0 ).I can try factoring this quadratic. Let me see, I need two numbers that multiply to -4 and add up to 3. Hmm, 4 and -1 work because 4 * (-1) = -4 and 4 + (-1) = 3.So, factoring:( (x + 4)(x - 1) = 0 )Therefore, the solutions are ( x = -4 ) and ( x = 1 ). So, the zeros are at ( x = -4 ) and ( x = 1 ).**Part II: Proving that ( b > 2 ) if the graph of ( f(x) ) is above the line ( y = x + 2 ).**Alright, so the graph of ( f(x) ) being above ( y = x + 2 ) means that for all real numbers ( x ), ( f(x) > x + 2 ).So, let's write that inequality:( ax^2 + (2a + 1)x + b > x + 2 )Subtract ( x + 2 ) from both sides:( ax^2 + (2a + 1)x + b - x - 2 > 0 )Simplify the terms:( ax^2 + 2a x + (b - 2) > 0 )So, we have ( ax^2 + 2a x + (b - 2) > 0 ) for all ( x ).This is a quadratic inequality. For a quadratic ( Ax^2 + Bx + C ) to be always positive, two conditions must hold:1. The leading coefficient ( A ) must be positive.2. The discriminant ( B^2 - 4AC ) must be negative (so that the quadratic doesn't cross the x-axis).Let's apply this to our quadratic ( ax^2 + 2a x + (b - 2) ).First, the leading coefficient is ( a ). So, ( a > 0 ).Second, the discriminant:( (2a)^2 - 4 * a * (b - 2) < 0 )Simplify:( 4a^2 - 4a(b - 2) < 0 )Factor out 4a:( 4a(a - (b - 2)) < 0 )Divide both sides by 4a (since ( a > 0 ), the inequality sign doesn't change):( a - (b - 2) < 0 )Simplify:( a - b + 2 < 0 )( -b + (a + 2) < 0 )( -b < -a - 2 )Multiply both sides by -1 (remember to flip the inequality sign):( b > a + 2 )Hmm, so ( b > a + 2 ). But wait, the problem states that the graph is above ( y = x + 2 ), so this must hold for all ( x ). But I need to find a condition solely on ( b ), not involving ( a ). Maybe I need another approach.Alternatively, since the inequality must hold for all ( x ), perhaps I can choose a specific value of ( x ) to simplify things. Let's choose ( x = 0 ):( f(0) = a*0 + (2a + 1)*0 + b = b )And the line ( y = x + 2 ) at ( x = 0 ) is ( y = 2 ).Since ( f(0) > 2 ), that gives ( b > 2 ).Oh, that's a simpler way! So, by evaluating at ( x = 0 ), we directly get ( b > 2 ).**Part III: Solving the inequality ( f(x) < 0 ) when ( b = 2 ).**So, with ( b = 2 ), the function becomes:( f(x) = ax^2 + (2a + 1)x + 2 )We need to solve ( ax^2 + (2a + 1)x + 2 < 0 ).First, let's factor this quadratic if possible.Looking for two numbers that multiply to ( a*2 = 2a ) and add up to ( 2a + 1 ). Hmm, maybe it's factorable.Let me try:( ax^2 + (2a + 1)x + 2 )Let me split the middle term:Looking for factors of ( a*2 = 2a ) that add up to ( 2a + 1 ). Hmm, perhaps ( 2a ) and 1.So, rewrite the middle term:( ax^2 + 2a x + x + 2 )Factor by grouping:( a x (x + 2) + 1(x + 2) )So, ( (a x + 1)(x + 2) )Yes, that works. So, ( f(x) = (a x + 1)(x + 2) )So, the inequality ( (a x + 1)(x + 2) < 0 ).To solve this inequality, we need to find the critical points where each factor is zero:1. ( a x + 1 = 0 ) ⇒ ( x = -1/a )2. ( x + 2 = 0 ) ⇒ ( x = -2 )So, the critical points are ( x = -2 ) and ( x = -1/a ).Now, the sign of the product depends on the intervals determined by these critical points. The intervals are:1. ( (-infty, -2) )2. ( (-2, -1/a) ) if ( -1/a > -2 )3. ( (-1/a, infty) )But the order of ( -2 ) and ( -1/a ) depends on the value of ( a ).Let me consider different cases for ( a ):**Case 1: ( a = 0 )**If ( a = 0 ), the function becomes:( f(x) = 0 + (0 + 1)x + 2 = x + 2 )So, the inequality ( x + 2 < 0 ) ⇒ ( x < -2 ).**Case 2: ( a neq 0 )**We need to consider the sign of ( a ) because it affects the direction of the inequality when multiplying or dividing.**Subcase 2.1: ( a > 0 )**If ( a > 0 ), then the parabola opens upwards.The critical points are ( x = -2 ) and ( x = -1/a ).We need to determine the order of ( -2 ) and ( -1/a ).If ( a > 0 ):- If ( a > 0 ), then ( -1/a ) is negative.- Compare ( -2 ) and ( -1/a ): - If ( a < 1/2 ), then ( -1/a < -2 ) because ( 1/a > 2 ). - If ( a = 1/2 ), then ( -1/a = -2 ). - If ( a > 1/2 ), then ( -1/a > -2 ).So, we have subcases:**Subcase 2.1.1: ( 0 < a < 1/2 )**Here, ( -1/a < -2 ). So, the critical points are ordered as ( -1/a < -2 ).The intervals are:1. ( (-infty, -1/a) )2. ( (-1/a, -2) )3. ( (-2, infty) )Since ( a > 0 ), the parabola opens upwards, so the product ( (a x + 1)(x + 2) ) is positive outside the roots and negative between them.But wait, actually, since it's a quadratic, the sign alternates between intervals.But since we have two roots, the sign will be:- Positive when ( x < -1/a )- Negative when ( -1/a < x < -2 )- Positive when ( x > -2 )But since ( -1/a < -2 ), the interval where the product is negative is ( (-1/a, -2) ).But wait, that seems counterintuitive because ( -1/a < -2 ), so between ( -1/a ) and ( -2 ), which is actually a smaller interval.But let's test with a specific value. Let me choose ( a = 1/3 ), so ( -1/a = -3 ).So, the critical points are ( x = -3 ) and ( x = -2 ).The intervals are:1. ( (-infty, -3) )2. ( (-3, -2) )3. ( (-2, infty) )Testing each interval:- For ( x < -3 ), say ( x = -4 ): ( (1/3*(-4) + 1) = (-4/3 + 1) = (-1/3) ), and ( (-4 + 2) = (-2) ). So, product is positive.- For ( -3 < x < -2 ), say ( x = -2.5 ): ( (1/3*(-2.5) + 1) = (-2.5/3 + 1) ≈ (-0.833 + 1) = 0.167 ), and ( (-2.5 + 2) = (-0.5) ). So, product is negative.- For ( x > -2 ), say ( x = 0 ): ( (1/3*0 + 1) = 1 ), and ( (0 + 2) = 2 ). Product is positive.So, the inequality ( (a x + 1)(x + 2) < 0 ) holds in ( (-3, -2) ), which is ( (-1/a, -2) ).**Subcase 2.1.2: ( a = 1/2 )**Here, ( -1/a = -2 ), so both roots coincide at ( x = -2 ).The quadratic becomes ( ( (1/2)x + 1 )(x + 2) ).But since both roots are the same, the quadratic touches the x-axis at ( x = -2 ).Since ( a > 0 ), the parabola opens upwards, so the quadratic is always non-negative, and only touches zero at ( x = -2 ). Therefore, the inequality ( f(x) < 0 ) has no solution.**Subcase 2.1.3: ( a > 1/2 )**Here, ( -1/a > -2 ). So, the critical points are ordered as ( -2 < -1/a ).The intervals are:1. ( (-infty, -2) )2. ( (-2, -1/a) )3. ( (-1/a, infty) )Testing the sign:- For ( x < -2 ), say ( x = -3 ): ( (a*(-3) + 1) ) and ( (-3 + 2) = (-1) ). Depending on ( a ), but since ( a > 1/2 ), ( a*(-3) + 1 ) could be negative or positive. Wait, let's pick a specific ( a ), say ( a = 1 ).So, ( f(x) = (x + 1)(x + 2) ).Testing intervals:- ( x < -2 ): say ( x = -3 ): ( (-3 + 1)(-3 + 2) = (-2)(-1) = 2 > 0 ).- ( -2 < x < -1 ): say ( x = -1.5 ): ( (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0 ).- ( x > -1 ): say ( x = 0 ): ( (0 + 1)(0 + 2) = 2 > 0 ).So, the inequality holds in ( (-2, -1) ), which is ( (-2, -1/a) ) since ( a = 1 ), ( -1/a = -1 ).Thus, in general, for ( a > 1/2 ), the solution is ( (-2, -1/a) ).**Subcase 2.2: ( a < 0 )**If ( a < 0 ), the parabola opens downwards.The critical points are ( x = -2 ) and ( x = -1/a ).Since ( a < 0 ), ( -1/a > 0 ).So, the critical points are ( x = -2 ) and ( x = -1/a ) where ( -1/a > 0 ).The intervals are:1. ( (-infty, -2) )2. ( (-2, -1/a) )3. ( (-1/a, infty) )Since the parabola opens downwards, the quadratic is positive between the roots and negative outside.So, the product ( (a x + 1)(x + 2) ) is:- Negative when ( x < -2 )- Positive when ( -2 < x < -1/a )- Negative when ( x > -1/a )Therefore, the inequality ( (a x + 1)(x + 2) < 0 ) holds when ( x < -2 ) or ( x > -1/a ).So, summarizing all cases:- If ( a < 0 ): ( x < -2 ) or ( x > -1/a )- If ( a = 0 ): ( x < -2 )- If ( 0 < a < 1/2 ): ( -1/a < x < -2 )- If ( a = 1/2 ): No solution- If ( a > 1/2 ): ( -2 < x < -1/a )That's a bit complex, but I think that's how it breaks down.**Final Answer**(I) The zeros are (boxed{-4}) and (boxed{1}).(II) It is proven that (b > 2).(III) The solution to the inequality depends on the value of (a):- If (a < 0), the solution is (boxed{{x mid x < -2 text{ or } x > -frac{1}{a}}}).- If (a = 0), the solution is (boxed{{x mid x < -2}}).- If (0 < a < frac{1}{2}), the solution is (boxed{{x mid -frac{1}{a} < x < -2}}).- If (a = frac{1}{2}), there is no solution, so (boxed{emptyset}).- If (a > frac{1}{2}), the solution is (boxed{{x mid -2 < x < -frac{1}{a}}}).