Cumulative Assessment

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Exercises 1 When we have the graph of an arithmetic function the n values are the number of the term and the an​ values are the term itself. Knowing this, we can interpret the given points as: (1,24)=(2,21)=(3,18)=(4,15)=​ First term is 24 Second term is 21 Third term is 18 Fourth term is 15​ Notice that the first term is 24 and each term is 3 less than the previous. This means that the arithmetic function has a1​=24 and d=-3. The only function that matches these requirements is option D.
Exercises 2 We have been given four inequalities and asked to determine which ones are equivalent. The easiest way to go about doing this is by simplifying all of them and comparing the results. Let's look at these one a time.First inequality To simplify the first inequality, we need to utilize the Subtraction Property of Inequalities. 3x+6≤8+2xLHS−2x≤RHS−2xx+6≤8LHS−6≤RHS−6x≤2Second inequality To simplify the second inequality, we need to remember that dividing by a negative requires us to reverse the direction of the inequality sign. 5x−5≥7x−9LHS+5≥RHS+55x≥7x−4LHS−7x≥RHS−7x-2x≥-4Divide by -2 and flip inequality signx≤2Third inequality To simplify the third inequality, we need to remember that dividing by a negative requires us to reverse the direction of the inequality sign. 12−3x≤18LHS−12≤RHS−12-3x≤6Divide by -3 and flip inequality signx≥-2Fourth inequality To simplify the fourth inequality, we can make it a bit easier if we multiply everything by -2 first. This will eliminate the fraction and all of the negatives. -2−23​x≥-3−xMultiply by -2 and flip inequality sign4+3x≤6+2xLHS−4≤RHS−43x≤2+2xLHS−2x≤RHS−2xx≤2Results Now we can compare the results. 3x+6≤8+2x5x−5≥7x−912−3x≤18-2−23​x≥-3−x​⇒x≤2⇒x≤2⇒x≥-2⇒x≤2​ The first, second, and fourth inequalities all have equivalent solution sets.
Exercises 3 If two variables are correlated, they either increase or decrease together. If the same two variables have a causal relationship, one of the variables must directly influence the other. Let's look at each of the given statements.a The first situation given is comparing the price of a pair of pants and the number sold. This situation has a negative correlation. The lower the price of a pair of pants, the more pairs that will be sold. If pants go on sale, more people will buy them. If more people can afford them, more will be sold. Also, high-end, over-priced, designer jeans usually have fewer pairs made knowing that fewer people will be able to buy them. Therefore, this is also a causal relationship.b Next, we are asked to compare the number of cell phones and the number of taxis in a city. As the population of a city gets bigger, the number of cell phones and the number of taxis within the city limits are likely to increase. These variables definitely have a strong, positive correlation but they do not have a causal relationship. The number of cell phones increases does not directly increase the number of taxis, the number of people within the city does.c The third situation is comparing a person's IQ and the time it takes the person to run 50 meters. There is no correlation or causation between these two variables. Highly intelligent people come in all shapes and sizes.d The last situation is comparing the amount of time spent studying and the score earned. The longer you study and prepare for an exam, the more likely you are to do well on that exam. Your score earned has a strong, positive correlation with the amount of time spent studying. This is also a causal relationship since one directly influences the other.Table SituationCorrelationCausation YesNoYesNo a.XXb.XX c.XX d.XX
Exercises 4 Let's begin by figuring out how each transformation affects the parent function, f(x)=x−1.Transformed Functionf(x)→?Simplified g(x)=f(x+2)(x+2)−1g(x)=x+1 k(x)=f(x) + 4(x−1) + 4k(x)=x+3 r(x)=3f(x)3(x−1)r(x)=3x−3 h(x)=f(3x)3x−1h(x)=3x−1 p(x)=f(-x)-x−1p(x)=-x−1 q(x)=-f(x)-(x−1)q(x)=-x+1 Now we can look at the given graph and find the slope and y-intercept for each function. We can then compare these requirements to the equations that we just found.FunctionColorSlopey-interceptEquationCorresponding function Blue-11y=-x+1q(x) Red3-1y=3x−1h(x) Green11y=x+1g(x) Purple13y=x+3k(x)
Exercises 5 We have been given a list of possible values for m and b for the slope-intercept form of our function. We need to choose which of these values are correct. First, let's use the given points, (6,1) and (-2,-3), to solve for the slope of the line. m=x2​−x1​y2​−y1​​Substitute (-2,-3) & (6,1)m=-2−6-3−1​Subtract termsm=-8-4​-b-a​=ba​m=21​ The slope of the line is 21​, so we have y=21​x+b. We can use either one of the given points to solve for the value of b. y=21​x+bx=6, y=11=21​⋅6+bMultiply1=3+bLHS−3=RHS−3-2=bRearrange equationb=-2 The final equation for the line is then: y=21​x−2, where m=21​ and b=-2.
Exercises 6 Notice that we have to determine whether the y-intercept of each part of the function is 3 or -3. The line defined for negative x-values intercepts the y-axis at y=3. Therefore, the first box should be filled with an addition sign. Since the circle at (0,3) is open, this line is not defined for x=0. The equation of the first piece is thus:We can tell that the second part of the function intercepts the y-axis at y=-3. We also know that it is defined for positive x-values including 0 because the circle at (0,-3) is closed. the equation of the second piece is thus:
Exercises 7 Let's choose the values to support our claim first, and then we will rearrange them to support our friend's claim.Our Claim For a relation to be function, each input value has to correspond with only one output value. Let's choose 5 different input values x, and 5 output values y.x-4-3-2-10 y01234 For each x-value, we have exactly one y-value. This is a function.Friend's claim For a relation not to be a function, at least one input value has to correspond with two or more output values. To create this, we can delete one of the inputs from our table and substitute it with another existing x-value. Let's substitute x=-4 with x=-3.x-3-3-2-10 y01234 For x=-3, there are two corresponding y-values. This is not a function.
Exercises 8 Let x be the price that would allow us to save the same amount of money regardless of which coupon we use. The first coupon takes 20% off our entire purchase. Using the percent proportion, ba​=100p​, we can find what "part" equals 20% of x by substituting b=x and p=20 into the formula and solving for a. ba​=100p​b=x, p=20xa​=10020​ Solve for a ba​=b/20a/20​xa​=51​LHS⋅x=RHS⋅xa=51​⋅xb1​⋅a=ba​ a=5x​ Using the first coupon, we save 5x​ dollars. The second coupon reduces our bill by a flat $5. To find x such that both coupons save us the same amount of money, we can equate the savings from each coupon and solve for x: 5x​=5⇔x=25 When our total bill at the restaurant is $25, both coupons will save us the same amount of money.
Exercises 9