Cumulative Assessment

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Exercises 1 To calculate the difference between the expected attendance, 65 people, and actual attendance, we need to subtract the expected attendance from the actual attendance. ∣actual−expected∣=deviation If we call the actual attendance x and substitute the given expected attendance, our expression becomes: ∣x−65∣=deviation. We are told that the actual attendance can vary by up to 30. Thus, we can write our equation as the following. ∣x−65∣=30 The equation corresponds to A.
Exercises 2
Exercises 3 To place each inequality in the correct box, we will solve each of them separately.5x−6+x≥2x−8 5x−6+x≥2x−8 Solve for x LHS+6≥RHS+65x+x≥2x−2Add terms6x≥2x−2LHS−2x≥RHS−2x4x≥-2LHS/4≥RHS/4x≥-42​ba​=b/2a/2​ x≥-21​ All values above -0.5 are solutions to the inequality. Therefore, any integer greater than this number is a solution, such as x=1 or x=2.2(3x+8)>3(2x+6) 2(3x+8)>3(2x+6) Solve for x Distribute 26x+16>3(2x+6)Distribute 36x+16>6x+18LHS−6x>RHS−6x 16≯18 Since 16 is never greater than 18, there is no solution to this inequality. This includes any possible integer solutions.17<4x+5<21 To solve this inequality, we have to separate it into two inequalities: 17<4x+5 and 4x+5<21. Now we solve each of them separately and then combine the solutions. 17<4x+5 Solve for x LHS−5<RHS−512<4xLHS/4<RHS/43<xRearrange inequality x>3 Let's solve the second inequality. 4x+5<21 Solve for x LHS−5<RHS−54x<16LHS/4<RHS/4 x<4 The final solution is 3<x<4. This means that x cannot be an integer because the solution set is all values between two consecutive integers.x−8+4x≤3(x−3)+2x x−8+4x≤3(x−3)+2x Solve for x Distribute 3x−8+4x≤3x−9+2xSimplify terms5x−8≤5x−9LHS−5x=RHS−5x-8≤-9Flip inequality and change signs 8≱9 Since 8 is not greater than or equal to 9, there are no solutions to this inequality. This includes any integer solutions.9x−3<12 or 6x+2>-10 We will solve each of the inequalities separately and then combine the solutions. 9x−3<12 Solve for x LHS+3<RHS+39x<15LHS/9<RHS/9x<915​Use a calculatorx<1.66666… x<1.67 So far we have that x<1.67. Since the original inequality is 9x−3<12 or 6x+2>-10, we can already tell that it has at least one integer solution. For example, x=1 satisfies the first inequality and this is enough.5(x−1)≤5x−3 5(x−1)≤5x−3 Solve for x Distribute 55x−5≤5x−3LHS−5x≤RHS−5x-5≤-3Flip inequality and change signs 5≥3 Because 5 is always greater than 3, we have reached an identity. The solution to this inequality is all real numbers. Therefore there is at least one integer solution.Conclusion Let's fill in the table with the information we have found.At least one integer solutionNo integer solutions 5x−6+x≥2x−89x−3<12 or 6x+2>-105(x−1)≤5x−32(3x+8)>3(2x+6) 17<4x+5<21 x−8+4x≤3(x−3)+2x
Exercises 4
Exercises 5 First, let's solve the given equation for x. 3(2x−4)=4(ax−2) Solve for x Distribute 36x−12=4(ax−2)Distribute 46x−12=4ax−8LHS+12=RHS+126x=4ax+4LHS−4ax=RHS−4ax6x−4ax=4Factor out xx(6−4a)=4LHS/4=RHS/4 x=6−4a4​ We can notice that the numerator is a positive number, 4. Because the fraction needs to simplify to be a positive number, we need the denominator to be positive as well. In other words, we must have that 6−4a>0. Let's solve this inequality for a. 6−4a>0 Solve for a LHS−6>RHS−6-4a>-6Flip inequality and change signs4a<6LHS/4<RHS/4a<46​ba​=b/2a/2​a<23​Write as a decimal a<1.5 When a<1.5, the solution of the equation is positive. From the given values, we identify the ones which are less than 1.5 as: -2,-1,0,1.
Exercises 6 Let's begin by determining the compound inequality that the solution set represents.Identifying the compound inequality Starting from the left, we see a closed circle at -2 which means x has to be greater than or equal to -2. We can write this as -2≤x. On the right, we have an open circle at 3 which means x has to be less than 3. This is written as x<3. Combining these inequalities, we have the following compound inequality x≥-2andx<3.Identifying the correct sign Next, we have to determine which signs we should fill the two gaps with. Since x is not isolated, we cannot fill them in right away. Let's begin by isolating x in both inequalities. 4x−18 22​ -x−3LHS+18 22​ RHS+184x 22​ -x+15LHS+x 22​ RHS+x5x 22​ 15LHS/5 22​ RHS/5x 22​ 3 Notice that we never divided or multiplied by a negative number. This means that the inequality sign stayed the same direction throughout solving it. Therefore, when comparing this to the compound inequality, we have: x 22​ 3⇒x <​ 3. Let's solve the next inequality. -3x−9 22​ -3LHS+3x 22​ RHS+3x-9 22​ -3+3xLHS+3 22​ RHS+3-6 22​ 3xLHS/3 22​ RHS/3-2 22​ x Again, notice that we did not divide or multiply by a negative number and the inequality sign stayed the same direction throughout. So when comparing this to the compound inequality, we get:
Exercises 7
Exercises 8