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Solve one-step and two-step equations: word problems

Puzzle Three\newlineSolve the following systems of equations. To unlock puzzle three find the product of each solution (ordered pair −x -x and y y value) and then find the mean of all five products.\newline11) −{3x+y=42x−y=6 -\left\{\begin{array}{l}3 x+y=4 \\ 2 x-y=6\end{array}\right. \newline22) −{2x+y=−33x+2y=−8 -\left\{\begin{array}{c}2 x+y=-3 \\ 3 x+2 y=-8\end{array}\right. \newline33) ⟶{7x−4y=−75x+y=22 \longrightarrow\left\{\begin{array}{c}7 x-4 y=-7 \\ 5 x+y=22\end{array}\right. \newline44) ⟶{−x+9y=−5x−5y=1 \longrightarrow\left\{\begin{aligned}-x+9 y & =-5 \\ x-5 y & =1\end{aligned}\right. \newline55) ⟶{x−y=−4−2x+3y=6 \longrightarrow\left\{\begin{array}{c}x-y=-4 \\ -2 x+3 y=6\end{array}\right. \newlineCode:
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Perweb Acoount\newline2020\newlineDL.\newlineHorne\newlineCPM Leaming Platf..\newlineHistory Overview\newlineMy IXL\newlineLearning\newlineAssessment\newlineM. 1919 Solve multi-step equations: complete the solution\newlinePGH\newlineComplete the process of solving the equation.\newlineFill in all missing terms and select all missing descriptions. Simplify any fra\newline\begin{tabular}{|c|c|c|c|}\newline\hline \multicolumn{22}{|c|}{8v+5+18v=14v+5 8 v+5+18 v=14 v+5 } & & \\\newline\hline 89v 89 \mathrm{v} & +5=14v+5 +5=14 v+5 & Combine like terms & 77 \\\newline\hline 75v 75 v & +5=5 +5=5 & Subtract 14v 14 v from both sides & : \\\newline\hline & 12v=0 12 v=0 & Subtract 55 from both sides & ∴ \therefore \\\newline\hline & v=0 v=0 & Divide both sides by 1212 & 11.11 \\\newline\hline\newline\end{tabular}\newlineSubmit
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USE THE TABLE BELOW TO HELP YOU FIND THE SURFACE AREA OF THE FIGURE BELOW. DRAG AND DROP THE MATCHING PIECE TO THE WHITE RECTANGLES.\newline\begin{tabular}{|c|c|c|c|}\newline\hline & SHAPE 11 & SHAPE 22 & SHAPE 33 \\\newline\hline FORMULA & & A=b⋅h2 A=\frac{b \cdot h}{2} & \\\newline\hline \begin{tabular}{l} \newlinePLUG IN \\\newlineVALUES\newline\end{tabular} & & & A=5⋅72 A=\frac{5 \cdot 7}{2} \\\newline\hline AREA & 2828 inches 2 ^{2} & & \\\newline\hline\newline\end{tabular}\newlineTOTAL SURFACE AREA\newline□ \square \newlineA=b⋅h2A=4⋅42A=5⋅72A=b⋅hA=4⋅7A=5⋅42 \begin{array}{l|l|l|} A=\frac{b \cdot h}{2} & A=\frac{4 \cdot 4}{2} & A=\frac{5 \cdot 7}{2} \\ \hdashline A=b \cdot h & A=4 \cdot 7 & A=\frac{5 \cdot 4}{2} \end{array} \newline3535 in 2 ^{2} \newlineA=5⋅42 A=\frac{5 \cdot 4}{2} \newline17.5in2 17.5 \mathrm{in}^{2} \newline1616 in 2 ^{2} \newlineA=4⋅7 A=4 \cdot 7 \newline88 in 2 ^{2} \newlineA=5⋅72 A=\frac{5 \cdot 7}{2} 00\newlineA=5⋅72 A=\frac{5 \cdot 7}{2} 11\newlineNotes
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Name:\newlineNARON\newlinePractice \& Problem Solving\newlineScan for\newlineMultimedia\newlineLeveled Practice In 1515−22-22, solve each equation.\newline1515.\newlinew−3.2=5.6w−3.2+□=5.6+□w=□ \begin{aligned} w-3.2 & =5.6 \\ w-3.2+\square & =5.6+\square \\ w & =\square \end{aligned} \newline(1717.)\newline48.55+k=61.7748.55+k−□=61.77−□k=□ \begin{aligned} 48.55+k & =61.77 \\ 48.55+k-\square & =61.77-\square \\ k & =\square \end{aligned} \newline1616. 9.6=1.6y 9.6=1.6 y \newline1818.\newline(1919.) 719=245+m 7 \frac{1}{9}=2 \frac{4}{5}+m \newline(2020.) a+314=529 a+3 \frac{1}{4}=5 \frac{2}{9} \newline(2121. 18⋅y=4 \frac{1}{8} \cdot y=4 \newline2222. x−638=467 x-6 \frac{3}{8}=4 \frac{6}{7} \newline(2323) Mr. Marlon buys these tickets for his family to visit the water park.\newlineThe total cost is $210 \$ 210 . Write and solve an equation to find the cost of each ticket.\newline2424. Higher Order Thinking Without solving, tell which equation below has a greater solution. Explain.\newline58m=23459m=234 \frac{5}{8} m=2 \frac{3}{4} \quad \frac{5}{9} m=2 \frac{3}{4} \newline2525. Mlake Sense and Persevere A high school track team's long jump record is 2121 feet 214 2 \frac{1}{4} inches. This year, Tim's best long jump is 2020 feet 912 9 \frac{1}{2} inches. If long jumps are measured to the nearest quarter inch, how much farther must Tim jump to break the record?\newline214214\newline44−5-5 Write and Solve Equations with Rational Numbers
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Question 22 of 66\newlineThis question: 11 point(s) possible\newlineA company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kWh860 \, \text{kWh}. You want to test this claim. You find that n=5n=5 residential customers has a mean monthly consumption of 890 kWh890\,\text{kWh}. Assume the population standard deviation is 128 kWh128\,\text{kWh}. At α=0.01\alpha=0.01, can you support the claim? Complete\newline(a) Identify H0H_{0} and HaH_{a}. Choose the correct answer below.\newlineA. H0:μ>860 (claim)H_{0}:\mu > 860 \, (\text{claim})\newlineB. H0:μ>890 (claim)H_{0}:\mu > 890 \, (\text{claim})\newlineHa:μ≤860H_{a}:\mu \leq 860\newlineC. n=5n=500\newlineD. n=5n=511\newlineE. n=5n=522\newlineF. n=5n=533\newline(Round to two decimal places as needed)\newlineA. The critical value is n=5n=544.\newlinea. The critical values are n=5n=555.\newlineIdentify the rejection region(s), Select the correct choice below.\newlineA. The rejection regions are n=5n=566 and n=5n=577.\newlineB. The rejection region is n=5n=577.\newlineC. The rejection region is n=5n=599
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Question 22 of 66\newlineThis question: 11 point(s) possible\newlineA company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kWh860 \, \text{kWh}. You want to test this claim. You find that a random sample of 6565 residential customers has a mean monthly consumption of 890 kWh890\,\text{kWh}. Assume the population standard deviation is 128 kWh128\,\text{kWh}. At α=0.01\alpha=0.01, can you support the claim? Complete parts (a) through ( 00 ).\newline(a) Identify H0H_{0} and HaH_{a}. Choose the correct answer below.\newlineA.\newline{H0:μ>860 (claim) H2:μ≤860\begin{cases} H_{0}:\mu > 860 \, \text{(claim)} \ H_{2}:\mu \leq 860 \end{cases}\newlineB.\newline{H0:μ>890 (claim) Ha:μ≤890\begin{cases} H_{0}:\mu > 890 \, \text{(claim)} \ H_{a}:\mu \leq 890 \end{cases}\newlineC.\newline{H0:μ≤890 Ha:μ>890 (claim)\begin{cases} H_{0}:\mu \leq 890 \ H_{a}:\mu > 890 \, \text{(claim)} \end{cases}\newlineD.\newline656500\newlineE.\newline656511\newlineF.\newline656522
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On a rainy day, Eva and her friends play a marble game called Revenge. To start, Eva divides a bag of marbles evenly among the 66 players. Each player gets 1313 marbles.\newlineWhich equation can you use to find the number of marbles mm in the bag?\newlineChoices:\newline(A) m6=13\frac{m}{6} = 13\newline(B) m+6=13m + 6 = 13\newline(C) m−6=13m - 6 = 13\newline(D) 6m=136m = 13\newlineSolve this equation for mm to find the number of marbles in the bag.\newline___\_\_\_ marbles\newline
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Using the Chi-Square Distribution Table, find the values for χleft2\chi_{\text{left}}^2 and χTight2\chi_{\text{Tight}}^2 of the following.\newline\newlineEspa\~{n}ol\newline\newlinePart 11 of 55\newline\newline(a) When α=0.01\alpha=0.01 and n=24n=24,\newline\newline\begin{cases}\(\newline\gamma_{\text{eft}}^2=9.260,(\newline\)\chi_{\text{tight}}^2=44.181\newline\end{cases}\)\newline\newlinePart: 15\frac{1}{5}\newline\newlinePart 22 of 55\newline\newline(b) When α=0.05\alpha=0.05 and n=29n=29,\newline\newline\begin{cases}\(\newline\chi_{\text{eft}}^2=,(\newline\)\chi_{\text{Tight}}^2=\newline\end{cases}\)\newline\newlineTry again\newline\newlineSkip Part\newline\newlineRecheck\newline\newlineSave For Later\newline\newlineSubmit Assignment\newline\newlineMacBook Air
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Based on the following calculator output, determine the population standard deviation of the dataset, rounding to the nearest 11ooth if necessary.\newline 1-Var-Stats xˉ=41.4285714286Σx=290Σx2=12714Sx=10.7990299388σx=9.99795897538n=7minX⁡=27Q1=31Med2=41Q3=53max⁡X=57 \begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=41.4285714286 \\ \Sigma x=290 \\ \Sigma x^{2}=12714 \\ S x=10.7990299388 \\ \sigma x=9.99795897538 \\ n=7 \\ \operatorname{minX}=27 \\ \mathrm{Q}_{1}=31 \\ \mathrm{Med}^{2}=41 \\ \mathrm{Q}_{3}=53 \\ \max \mathrm{X}=57 \end{array} \newlineAnswer:
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Urpi was given this problem:\newlineThe side b(t) b(t) of the base of a square prism is decreasing at a rate of 77 kilometers per minute and the height h(t) h(t) of the prism is increasing at a rate of 1010 kilometers per minute. At a certain instant t0 t_{0} , the base's side is 44 kilometers and the height is 99 kilometers. What is the rate of change of the surface area S(t) S(t) of the prism at that instant?\newlineWhich equation should Urpi use to solve the problem?\newlineChoose 11 answer:\newline(A) S(t)=6[b(t)]2 S(t)=6[b(t)]^{2} \newline(B) S(t)=[b(t)]3 S(t)=[b(t)]^{3} \newline(C) S(t)=2[b(t)]2+4b(t)⋅h(t) S(t)=2[b(t)]^{2}+4 b(t) \cdot h(t) \newline(D) S(t)=[b(t)]2⋅h(t) S(t)=[b(t)]^{2} \cdot h(t)
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