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One-step inequalities: word problems

Terry was asked to determine whether f(x)=x3+1x f(x)=x^{3}+\frac{1}{x} is even, odd, or neither. Here is his work:\newlineStep 11: Find expression for f(−x) f(-x) \newlinef(−x)=(−x)3+1(−x)=−x3−1x \begin{aligned} f(-x) & =(-x)^{3}+\frac{1}{(-x)} \\ & =-x^{3}-\frac{1}{x} \end{aligned} \newlineStep 22: Check if f(−x) f(-x) is equal to f(x) f(x) or −f(x) -f(x) \newline−x3−1x -x^{3}-\frac{1}{x} isn't the same as f(x)=x3+1x f(x)=x^{3}+\frac{1}{x} or\newline−f(x)=−x3+1x.  -f(x)=-x^{3}+\frac{1}{x} \text {. } \newlineStep 33: Conclusion\newlinef(−x) f(-x) isn't equivalent to either f(x) f(x) or −f(x) -f(x) , so f f is neither even nor odd.\newlineIs Terry's work correct? If not, what is the first step where Terry made a mistake?\newlineChoose 11 answer:\newline(A) Terry's work is correct.\newline(B) Terry's work is incorrect. He first made a mistake in Step 11.\newline(C) Terry's work is incorrect. He first made a mistake in Step 22.\newline(D) Terry's work is incorrect. He first made a mistake in Step 33.
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Tom was asked to determine whether f(x)=x3+x f(x)=x^{3}+x is even, odd, or neither. Here is his work:\newlineStep 11: Find expression for f(−x) f(-x) \newlinef(−x)=(−x)3+x=−x3+x \begin{aligned} f(-x) & =(-x)^{3}+x \\ & =-x^{3}+x \end{aligned} \newlineStep 22: Check if f(−x) f(-x) is equal to f(x) f(x) or −f(x) -f(x) \newline−x3+x -x^{3}+x isn't the same as f(x)=x3+x f(x)=x^{3}+x or −f(x)=−x3−x -f(x)=-x^{3}-x .\newlineStep 33: Conclusion\newlinef(−x) f(-x) isn't equivalent to either f(x) f(x) or −f(x) -f(x) , so f f is neither even nor odd.\newlineIs Tom's work correct? If not, what is the first step where Tom made a mistake?\newlineChoose 11 answer:\newline(A) Tom's work is correct.\newline(B) Tom's work is incorrect. He first made a mistake in Step 11.\newline(C) Tom's work is incorrect. He first made a mistake in Step 22.\newline(D) Tom's work is incorrect. He first made a mistake in Step 33.
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Jen was asked to determine whether f(x)=1x3 f(x)=\frac{1}{\sqrt[3]{x}} is even, odd, or neither. Here is her work:\newlineStep 11: Find expression for f(−x) f(-x) \newlinef(−x)=1(−x)3=1−13⋅x3=−1x3 \begin{aligned} f(-x) & =\frac{1}{\sqrt[3]{(-x)}} \\ & =\frac{1}{\sqrt[3]{-1} \cdot \sqrt[3]{x}} \\ & =-\frac{1}{\sqrt[3]{x}} \end{aligned} \newlineStep 22: Check if f(−x) f(-x) is equal to f(x) f(x) or −f(x) -f(x) \newline−1x3 -\frac{1}{\sqrt[3]{x}} is the same as\newline−f(x)=−1x3. -f(x)=-\frac{1}{\sqrt[3]{x}} . \newlineStep 33: Conclusion\newlinef(−x) f(-x) is equivalent to −f(x) -f(x) , so f f is even.\newlineIs Jen's work correct? If not, what is the first step where Jen made a mistake?\newlineChoose 11 answer:\newline(A) Jen's work is correct.\newline(B) Jen's work is incorrect. She first made a mistake in Step 11.\newline(C) Jen's work is incorrect. She first made a mistake in Step 22.\newline(D) Jen's work is incorrect. She first made a mistake in Step 33.
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April was asked to determine whether f(x)=x3−1 f(x)=x^{3}-1 is even, odd, or neither. Here is her work:\newlineStep 11: Find expression for f(−x) f(-x) \newlinef(−x)=(−x)3−1=−x3−1 \begin{aligned} f(-x) & =(-x)^{3}-1 \\ & =-x^{3}-1 \end{aligned} \newlineStep 22: Check if f(−x) f(-x) is equal to f(x) f(x) or −f(x) -f(x) \newline−x3−1 -x^{3}-1 is the same as\newline−f(x)=−x3−1 -f(x)=-x^{3}-1 .\newlineStep 33: Conclusion\newlinef(−x) f(-x) is equivalent to −f(x) -f(x) , so f f is odd.\newlineIs April's work correct? If not, what is the first step where April made a mistake?\newlineChoose 11 answer:\newline(A) April's work is correct.\newline(B) April's work is incorrect. She first made a mistake in Step 11.\newline(C) April's work is incorrect. She first made a mistake in Step 22.\newline(D) April's work is incorrect. She first made a mistake in Step 33.
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Kris was asked to determine whether f(x)=x3−x f(x)=x^{3}-x is even, odd, or neither. Here is his work:\newlineStep 11: Find expression for f(−x) f(-x) \newlinef(−x)=(−x)3−(−x)=(−1)3⋅x3+x=−x3+x \begin{aligned} f(-x) & =(-x)^{3}-(-x) \\ & =(-1)^{3} \cdot x^{3}+x \\ & =-x^{3}+x \end{aligned} \newlineStep 22: Check if f(−x) f(-x) is equal to f(x) f(x) or −f(x) -f(x) \newline−x3+x -x^{3}+x is the same as\newline−f(x)=−x3+x.  -f(x)=-x^{3}+x \text {. } \newlineStep 33: Conclusion\newlinef(−x) f(-x) is equivalent to −f(x) -f(x) , so f f is odd.\newlineIs Kris' work correct? If not, what is the first step where Kris made a mistake?\newlineChoose 11 answer:\newline(A) Kris' work is correct.\newline(B) Kris' work is incorrect. He first made a mistake in Step 11.\newline(C) Kris' work is incorrect. He first made a mistake in Step 22.\newline(D) Kris' work is incorrect. He first made a mistake in Step 33.
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Sydney was asked whether the following equation is an identity:\newline(2x2−18)(x+3)(x−3)=2x4−36x2+162 \left(2 x^{2}-18\right)(x+3)(x-3)=2 x^{4}-36 x^{2}+162 \newlineShe performed the following steps:\newline(2x2−18)(x+3)(x−3) \left(2 x^{2}-18\right)(x+3)(x-3) \newline↪ Step 1=2(x2−9)(x2−3x+3x−9) \stackrel{\text { Step } 1}{\hookrightarrow}=2\left(x^{2}-9\right)\left(x^{2}-3 x+3 x-9\right) \newline↪ Step 2=2(x2−9)(x2−9) \stackrel{\text { Step } 2}{\hookrightarrow}=2\left(x^{2}-9\right)\left(x^{2}-9\right) \newline↪ Step 3=2(x2−9)2 \stackrel{\text { Step } 3}{\hookrightarrow}=2\left(x^{2}-9\right)^{2} \newline↪ Step 4=2(x4−18x2+81) \stackrel{\text { Step } 4}{\hookrightarrow}=2\left(x^{4}-18 x^{2}+81\right) \newline↪ Step 5=2x4−36x2+162 \stackrel{\text { Step } 5}{\hookrightarrow}=2 x^{4}-36 x^{2}+162 \newlineFor this reason, Sydney stated that the equation is a true identity.\newlineIs Sydney correct? If not, in which step did she make a mistake?\newlineChoose 11 answer:\newline(A) Sydney is correct.\newline(B) Sydney is incorrect. She made a mistake in step 11.\newline(C) Sydney is incorrect. She made a mistake in step 33.\newline(D) Sydney is incorrect. She made a mistake in step 44.
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Grade 88 Pre-Algebra\newlineMA.88.DP.11.11 Practice\newlineName: \qquad Date: +15/24 +15 / 24 \newline11. The table below shows the number of people and the cost for admission into various fairs throughout the state of Florida.\newlineFair Admission Prices throughout Florida\newline\begin{tabular}{|c|c|}\newline\hline Number of People & Admission Cost \\\newline\hline 22 & $30 \$ 30 \\\newline\hline 33 & $45 \$ 45 \\\newline\hline 33 & $42 \$ 42 \\\newline\hline 44 & $80 \$ 80 \\\newline\hline 55 & $160 \$ 160 \\\newline\hline 66 & $90 \$ 90 \\\newline\hline\newline\end{tabular}\newlineSelect the type of graph and explanation that is appropriate for the fair admissions data throughout the state of Florida.\newline(11) line graph\newlineA\newline(B) scatter plot is appropriate because it shows\newlineC) how the prices increase as the number of people attending increases.\newline(66) the data points illustrate a pattern.\newline() the data is continuous.
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1717. Nebula used black and white square tiles to form patterns as shown below. She recorded the number of black and white tiles used in the table.\newlinePattern 11\newlinePattern 22\newlinePattern 33\newline\begin{tabular}{|c|c|c|c|}\newline\hline & \begin{tabular}{c} \newlineTotal Number of \\\newlinetiles\newline\end{tabular} & \begin{tabular}{c} \newlineNumber of Black \\\newlinetiles\newline\end{tabular} & \begin{tabular}{c} \newlineNumber of White \\\newlinetiles\newline\end{tabular} \\\newline\hline Pattern 11 & 33 & 22 & 11 \\\newline\hline Pattern 22 & 66 & 33 & 33 \\\newline\hline Pattern 33 & 99 & 55 & 44 \\\newline\hline Pattern 44 & 1212 & 66 & 66 \\\newline\hline\newline\end{tabular}\newline(a) Complete the table above for Pattern 44. [11]\newline(b) How many black tiles did Nebula use in Pattern 8585 ?
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P66 Math - Fraction Ex. 11\newline44. Kezia used 512 \frac{5}{12} of her money to buy some pens and 67 \frac{6}{7} of the remainder to buy f f 33 files. A file costs 44 times as much as a pen. How many pens did she buy?\newlineleft 712 \frac{7}{12} \newline33 files: 12 \frac{1}{2} \newlinePen: 12×13×14×124 \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{24} \qquad \newline2828 children shared a packet of sweets equally. When 44 children gave up 34 \frac{3}{4} of their share, the remaining children received 22 more sweets each. How many sweets were there in the packet at first?\newline28+4=2424×2=4848−34×4=48−3=4545+48=93 \begin{array}{l} 28+4=24 \\ 24 \times 2=48 \\ 48-\frac{3}{4} \times 4=48-3=45 \\ 45+48=93 \end{array} \newlineAns: \qquad \newlineThere were 120120 apples in basket A and basket B. 23 \frac{2}{3} of the apples in basket A\newlinewere red and the rest were greent. 67 \frac{6}{7} 00 of the apples in basket 67 \frac{6}{7} 11 were red and the rest were green. The number of green apples in both baskets was equal. How many red apples were there altogether?
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You wish to test the following claim (Ha) \left(H_{a}\right) at a significance level of α=0.10 \alpha=0.10 .\newlineHo:μ=60.8Ha:μ>60.8 \begin{array}{l} H_{o}: \mu=60.8 \\ H_{a}: \mu>60.8 \end{array} \newlineYou believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:\newline\begin{tabular}{|r|}\newline\hline data \\\newline\hline 6464.88 \\\newline\hline 3333.66 \\\newline\hline 6969.66 \\\newline\hline 5454.77 \\\newline\hline 7979 \\\newline\hline 6969.66 \\\newline\hline 8686.99 \\\newline\hline 6969.66 \\\newline\hline 9393.55 \\\newline\hline 9999.22 \\\newline\hline 9090.66 \\\newline\hline 7676.77 \\\newline\hline 6565.88 \\\newline\hline\newline\end{tabular}\newlineWhat is the critical value for this test? (Report answer accurate to three decimal places.)\newlinecritical value =1.356 =1.356 0∘ 0^{\circ} \newlineWhat is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = = □ \square
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You wish to test the following claim (Ha) \left(H_{a}\right) at a significance level of α=0.10 \alpha=0.10 .\newlineHo:μ=60.8Ha:μ>60.8 \begin{array}{l} H_{o}: \mu=60.8 \\ H_{a}: \mu>60.8 \end{array} \newlineYou believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:\newline\begin{tabular}{|r|}\newline\hline data \\\newline\hline 6464.88 \\\newline\hline 3333.66 \\\newline\hline 6969.66 \\\newline\hline 5454.77 \\\newline\hline 7979 \\\newline\hline 6969.66 \\\newline\hline 8686.99 \\\newline\hline 6969.66 \\\newline\hline 9393.55 \\\newline\hline 9999.22 \\\newline\hline 9090.66 \\\newline\hline 7676.77 \\\newline\hline 6565.88 \\\newline\hline\newline\end{tabular}\newlineWhat is the critical value for this test? (Report answer accurate to three decimal places.)\newlinecritical value = = □ \square \newlineWhat is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = = □ \square
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Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 2929 people who drank ethanol and another group of 2929 people given a placebo. The errors for the treatment group have a standard deviation of 22.1010 , and the errors for the placebo group have a standard deviation of 00.8282 . Assume that the two populations are normally distributed. Use a 00.0505 significance level to test the claim that both groups have the same amount of variation among th errors.\newlineLet sample 11 be the sample with the larger sample variance, and let sample 22 be the sample with the smalle sample variance. What are the null and alternative hypotheses?\newlineA. H0:σ12≠σ22 H_{0}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \newlineB. H0:σ12=σ22 H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \newlineH1:σ12=σ22H1:σ12>σ22 \mathrm{H}_{1}: \sigma_{1}^{2}=\sigma_{2}^{2} \quad \mathrm{H}_{1}: \sigma_{1}^{2}>\sigma_{2}^{2} \newlineC. H0:σ12=σ22 H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \newlineD.\newlineH0:σ12=σ22H1:σ12≠σ22 \begin{array}{l} H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \\ H_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2} \end{array} \newlineH1:σ12<σ22 \mathrm{H}_{1}: \sigma_{1}^{2}<\sigma_{2}^{2} \newlineIdentify the test statistic.\newline□ \square (Round to two decimal places as needed.)
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QUIZ/ASSESSMENT NO: 44.11\newlineUse triangle inequality theorems to answer the following activity.\newlineA. Write the inequality relating to the given pair of segments or angles below. Refer to the figure on the right.\newline11. LO‾_LV‾ \overline{L O} \_\overline{L V} \newline22. VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} \newline33. m∠YOE m \angle Y O E \qquad m∠OYE m \angle O Y E \newline44. m∠LOV m \angle L O V \qquad m∠LVO m \angle L V O \newlineB. Determine if the given lengths of a triangle can be the measure of its sides by writing YES or NO after each given.\newline11. 8,10,11 8,10,11 \newline33. 3,6,9 3,6,9 \newlineVY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 00\newline22. 44, 99, 1616\newline44. VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 11\newlineC. Write the inequality relating to the given pair of angles on the right.\newline11. VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 22 \qquad VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 44\newline22. VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 44 \qquad VY‾⟶OY‾ \overline{V Y} \longrightarrow \overline{O Y} 77
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CHAPTER 88 / BASIC ALGEBRA\newlinePRACTICE SAT QUESTIONS\newline11. If y=4 y=4 and x=3 x=3 , then y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= \newlineA. 2424\newlineB. 2525\newlineC. 2626\newlineD. 2727\newlineE. 2828\newline22. If 32+b=5 3^{2}+b=5 , then a2+2ab+b2= a^{2}+2 a b+b^{2}= \newline77. Which of the following statements is true?\newlineI. 4x2−9y2=(2x−3y)2 4 x^{2}-9 y^{2}=(2 x-3 y)^{2} .\newlineII. (x−4)2=x2−16 (x-4)^{2}=x^{2}-16 .\newlineIII. x2+2xy+y2=(x+y)2 x^{2}+2 x y+y^{2}=(x+y)^{2} .\newlineA. I\newlineB. II\newlineC. III\newlineD. I and II\newlineE. I and III\newlineA. 55\newlineB. 1010\newline33\newlineC. 1515\newlineD. 2020\newline(E.) 2525\newline9+12+4 9+12+4 \newline33. x=x2−5x+6,a=2,a= x=x^{2}-5 x+6, a=2, a= \newlineA. −4-4\newlineB. 44\newlineC. −4-4 and 44\newlineD. 11 and 44\newlineE. 11\newline44. x=3 x=3 00\newlineA. 00 and 33\newlineB. 33\newlineC. 66\newlineD. 00 and 66\newlineE. 33 and 66\newline55. x=3 x=3 11\newlineA. x=3 x=3 22\newlineB. x=3 x=3 33\newlineC. x=3 x=3 44\newlineD. x=3 x=3 55\newlineE. x=3 x=3 66\newline66. x=3 x=3 77. What is the sum of the possible value of x=3 x=3 88 ?\newlineA. 33\newlineB. 33.55\newlineC. 44\newlineD. 44.55\newlineE. 55\newline88. x=3 x=3 99\newlineA. 11\newlineB. −1-1\newlineC. 00\newlineD. 22\newlineE. −2-2\newline99. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 00\newlineA. 22\newlineB. 33\newlineC. 44\newlineD. 66\newlineE. 1212\newline1010. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 11\newlineA. −65-65\newlineB. −50-50\newlineC. −35-35\newlineD. −20-20\newlineE. −15-15\newline1111. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 22\newlineA. 8181\newlineB. 99\newlineC. 144144\newlineD. 1212\newlineE. 169169\newline1212. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 33\newlineA. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 44\newlineB. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 55\newlineC. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 66\newlineD. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 77\newlineE. y3−3x2+3y−2xy+x= y^{3}-3 x^{2}+3 y-2 x y+x= 88
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Socially 00ptimal Level\newlineHow does this compare to the socially optimal level of provision? The social optimum is the quantity at which the sum of the individuals' marginal rates of substitution equals the ratio of prices (which is 11 in this example). Each individual's MRS M R S is the ratio of his marginal utility of fireworks to his marginal utility of private goods, which we obtain by differentiating the previous utility function with respect to fireworks and then again with respect to private goods. So the optimal amount of fireworks is determined by:\newline(100−FB)/[2×(FB+FJ)]+(100−Fj)/[2×(FB+Fj)]=1 \left(100-F_{B}\right) /\left[2 \times\left(F_{B}+F_{J}\right)\right]+\left(100-F_{j}\right) /\left[2 \times\left(F_{B}+F_{j}\right)\right]=1 \newlineUsing the fact that total fireworks F=FB+Fβ F=F_{B}+F_{\beta} we can rewrite this equation as:\newline(200−F)/2F=1 (200-F) / 2 F=1 \newlineSolving this, we obtain F=66.6 F=66.6 . This quantity is much higher than the total provision by the private market, 4040 , due to the free rider problem. The public good is underprovided by the private market.
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A credit card company claims that the mean credit card debt for individuals is greater than $5,100\$5,100. You want to test this claim. You find that a random sample of 3333 cardholders has a mean credit card balance of $5,301\$5,301 and a standard deviation of $575\$575. At α=0.01\alpha=0.01, can you support the claim? Complete parts (a) through (e) below. Assume the population is normally distributed. (a) Write the claim mathematically and identily H0H_{0} and H2H_{2}. Which of the following correctly states H0H_{0} and H2H_{2} ?\newlineA.\newlineH0:μ=$5,100 H2:μ≠$5,100\begin{array}{l} H_{0}:\mu=\$5,100 \ H_{2}:\mu\neq\$5,100 \end{array}\newlineB.\newlineH0:μ≥$5,100 H2:μ<$5,100\begin{array}{l} H_{0}:\mu \geq \$5,100 \ H_{2}:\mu < \$5,100 \end{array}\newlineC.\newline$5,301\$5,30100\newlineD.\newline$5,301\$5,30111\newlineE.\newline$5,301\$5,30122\newlineF.\newline$5,301\$5,30133\newline(B) Find the critical value(s) and identify the rejection region(s). What is(are) the critical value(s), $5,301\$5,30144?\newline$5,301\$5,30155
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A company claims that the mean monthly residential electricity consumption in a certain region is more than 860kiloWatt−hours(kWh). 860 \mathrm{kiloWatt-hours} \mathrm{(kWh).} You want to test this claim. You find that a random sample of 6565 residential customers has a mean monthly consumption of 890kWh 890 \mathrm{kWh} . Assume the population standard deviation is 128kWh 128 \mathrm{kWh} . At α=0.01 \alpha=0.01 , can you support the claim? Complete parts (a) through ( 00 ).\newline(a) Identify H0 \mathrm{H}_{0} and Ha \mathrm{H}_{\mathrm{a}} . Choose the correct answer below.\newlineA. H0:μ>860 (claim) H8:μ≤860 \begin{array}{l} H_{0}: \mu>860 \text { (claim) } \\ H_{8}: \mu \leq 860 \end{array} \newlineB. H0:μ>890 (claim) Ha:μ≤890 \begin{array}{l} H_{0}: \mu>890 \text { (claim) } \\ H_{a}: \mu \leq 890 \end{array} \newlinec. H0:μ≤890H2:μ>890 (claim)  \begin{array}{l} H_{0}: \mu \leq 890 \\ H_{2}: \mu>890 \text { (claim) } \end{array} \newlineD. H0:μ=860 (claim) Ha:μ≠860 \begin{array}{l} H_{0}: \mu=860 \text { (claim) } \\ H_{a}: \mu \neq 860 \end{array} \newlineE. H0:μ=890H8:μ≠890 (daim)  \begin{array}{l} H_{0}: \mu=890 \\ H_{8}: \mu \neq 890 \text { (daim) } \end{array} \newlineF. H0:μ≤860Ha:μ>860 (claim)  \begin{array}{l} H_{0}: \mu \leq 860 \\ H_{a}: \mu>860 \text { (claim) } \end{array}
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blem 44 .\newlineblem 55 ...\newlineblem 6✓ 6 \checkmark \newlineablem 7 7 \sqrt{ } \newlineoblem 8 8 \sqrt{ } \newlineoblem 9 9 \sqrt{ } \newlineoblem 1010 v\newlineAt least one of the answers above is NOT correct.\newline(a) Write the equation ∣6−T3∣=5 \left|6-\frac{T}{3}\right|=5 as two separate equations, and enter each equation in its own answer box below. Neither of your equations should use absolute value.\newline3=0 help (equations)  help (equations)  \begin{array}{ll} 3=0 & \text { help (equations) } \\ \text { help (equations) } \end{array} \newline(b) Solve both equations above, and enter your answers as a comma separated list.\newliner=3,33 help (numbers)  r=3,33 \quad \text { help (numbers) }
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