5.3 使用线性模型的应用

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  使用线性模型的应用

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  Applications using Linear Models 
  ::使用线性模型的应用

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  S olve word problems using  the equation of a straight line.
  ::使用直线的方程式解决字问题 。

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  Real-World Application: Moving Trucks 
  ::真实世界应用程序:移动卡车

  

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  Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some number of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges? Write an equation in point- form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?
  ::Marciel租了一辆流动卡车。 Marciel只记得租来的卡车公司每天收费40美元,每英里约几美分。 Marciel开的是46英里,账单的最后金额(税前)是63美元。卡车租赁公司每英里收费的金额是多少? 写一个方程式来描述这种情况。如果Marciel开的是220英里,租这辆卡车要花多少钱?

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  Let’s define our variables:
  ::让我们定义我们的变数:

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  $$\begin{array}{rl}x& =\text{distance in miles}\\ y& =\text{cost of the rental truck}\end{array}$$

  
  ::x=英里距离=租车成本

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  Marciel pays a flat fee of $40 for the day; this is the $\begin{array}{r}y\end{array}$ - intercept .
  ::Marciel每天支付40美元的定额费用,这是Y调查。

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  He pays $63 for 46 miles; this is the coordinate point (46,63).
  ::他付63美元,46英里;这是坐标点(46,63)。

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  Start with the point-slope form of the line: $\begin{array}{r}y-y_0=m(x-x_0)\end{array}$
  ::以线条的点斜体形式开始: y- y0=m( x- x0)

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  Plug in the coordinate point: $\begin{array}{r}63-y_0=m(46-x_0)\end{array}$
  ::坐标点中的插件: 63- y0=m( 46- x0)

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  Plug in the point (0, 40): $\begin{array}{r}63-40=m(46-0)\end{array}$
  ::插插点(0,40):63-40=m(46-0)

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  Solve for the slope: $\begin{array}{r}23=46m\to m=\frac{23}{46}=0.5\end{array}$
  ::坡度: 23= 46mm=2346=0. 5

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  The slope is 0.5, so the truck company charges 0.5 dollars, or 50 cents, per mile ($0.5 = 50 cents). Plugging in the slope and the $\begin{array}{r}y\end{array}$ -intercept, the equation of the line is $\begin{array}{r}y=0.5x+40\end{array}$ .
  ::斜坡为0.5,所以卡车公司每英里收费0.5美元或50美分(0.5美元=50美分),在斜坡和y-intercut中插入,线的方程式是y=0.5x+40。

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  To find out the cost of driving the truck 220 miles, we substitute 220 for  $\begin{array}{r}x\end{array}$  to get $\begin{array}{r}y-40=0.5(220)\Rightarrow y=\mathrm{\$}150\end{array}$ .
  ::为了了解卡车220英里的驾驶费用,我们用220英里代替xx,以获得y-40=0.5(220)y=150美元。

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  Driving 220 miles would cost $150.
  ::开车220英里需要150美元

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  Real-World Application: Sales Commission 
  ::实际世界应用:销售委员会

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  Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month, she adds up sales and she figures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. Use the equation to determine  Anne’s monthly base salary.
  ::Anne得到了一份销售窗口窗帘的工作。 她每卖一个窗口窗帘都得到每月基本工资和6美元的佣金。 月底,她加了销售额,发现她卖了200个窗口窗帘,赚了2500美元。 用点窗帘形式写一个方程式来描述这种情况。 用方程式来确定Anne的月基本工资。

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  First  define the variables:
  ::首先定义变量 :

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  $$\begin{array}{rl}x& =\text{number of window shades sold}\\ y& =\text{Anne's earnings}\end{array}$$

  
  ::x = 售出的窗帘数= Anne的收入

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  You  are given the slope and a point on the line:
  ::给您一个斜坡和线上的一个点 :

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  Nadia gets $6 for each shade, so the slope is 6.
  ::Nadia每个阴影6美元,所以斜坡是6美元。

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  She made $2500 when she sold 200 shades, so there is a point at (200, 2500).
  ::她卖了200个窗帘,赚了2500美元 所以在200 2500分时有个点

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  Start with the point-slope form of the line: $\begin{array}{r}y-y_0=m(x-x_0).\end{array}$
  ::开始于线条的点窗体: y- y0=m( x- x0) 。

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  Plug in the slope: $\begin{array}{r}y-y_0=6(x-x_0).\end{array}$
  ::斜坡中的插件:y-y0=6(x-x0)。

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  Plug in the point (200, 2500): $\begin{array}{r}y-2500=6(x-200).\end{array}$
  ::插插点 (200, 2500): y-2500=6(x- 200) 。

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  To find Anne’s base salary, we plug in $\begin{array}{r}x=0\end{array}$ and get $\begin{array}{r}y-2500=-1200\Rightarrow y=\mathrm{\$}1300.\end{array}$
  ::为了找到安妮的基薪,我们插入x=0,并获得y-25001200y=1300美元。

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  Anne’s monthly base salary is $1300.
  ::安妮月基薪为1300美元。

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  Real-World Application: Buying Fruit 
  ::真实世界应用程序:购买水果

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  Nadia buys fruit at her local farmer’s market. This Saturday, oranges cost $2 per pound and cherries cost $3 per pound. She has $12 to spend on fruit. Write an equation in standard form that describes this situation. If she buys 4 pounds of oranges, how many pounds of cherries can she buy?
  ::Nadia在当地农民市场购买水果。 本周六,橙子每磅2美元,樱桃每磅3美元。 她有12美元用于水果。 她有12美元用于水果。 用标准格式写一个方程式描述这种情况。 如果她购买4磅橙子,她能买多少樱桃?

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  Let’s define our variables:
  ::让我们定义我们的变数:

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  $$\begin{array}{rl}x& =\text{pounds of oranges}\\ y& =\text{pounds of cherries}\end{array}$$

  
  ::x=千吨橘子=磅樱桃

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  The equation that describes this situation is $\begin{array}{r}2x+3y=12.\end{array}$
  ::描述这种情况的方程式是 2x+3y=12。

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  If she buys 4 pounds of oranges, we can plug $\begin{array}{r}x=4\end{array}$ into the equation and solve for $\begin{array}{r}y\end{array}$ :
  ::如果她买4磅橙子 我们可以在方程式中插入 x=4 并解决y:

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  $$\begin{array}{rl}2(4)+3y& =12\\ 3y& =12-8\\ 3y& =4\\ y& =\frac{4}{3}\end{array}$$

  
  ::2(4)+3y=123y=12-83y=4y=43

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  Nadia can buy $\begin{array}{r}1\frac{1}{3}\end{array}$ pounds of cherries.
  ::Nadia可以买113磅樱桃

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  Example
  ::示例示例示例示例

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  Example 1
  ::例1

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  Peter skateboards part of the way to school and walks the rest of the way. He can skateboard at 7 miles per hour and he can walk at 3 miles per hour. The distance to school is 6 miles. Write an equation in standard form that describes this situation. If he skateboards for $\begin{array}{r}\frac{1}{2}\end{array}$ an hour, how long does he need to walk to get to school?
  ::彼得滑板是上学路段的一部分,走剩下的路。他每小时可以滑板7英里,每小时可以行走3英里。上学的距离是6英里。用标准格式写一个方程,描述这种情况。如果他滑板每小时12小时,他需要步行多久才能上学?

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  D efine the  variables:
  ::定义变量 :

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  $$\begin{array}{rl}x& =\text{time Peter skateboards}\\ y& =\text{time Peter walks}\end{array}$$

  
  ::彼得滑板滑板滑板滑板滑雪

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  The equation that describes this situation is: $\begin{array}{r}7x+3y=6.\end{array}$
  ::描述这种情况的方程式是:7x+3y=6。

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  Peter skateboards $\begin{array}{r}\frac{1}{2}\end{array}$ an hour, so substitute   $\begin{array}{r}x=0.5\end{array}$ into the equation and solve for $\begin{array}{r}y:\end{array}$
  ::彼得滑板每小时12小时, 替换x=0. 5的方程式, 并解决y:

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  $$\begin{array}{rl}7(0.5)+3y& =6\\ 3y& =6-3.5\\ 3y& =2.5\\ y& =\frac{5}{6}\end{array}$$

  
  ::7(0.5)+3y=63y=6-3.53y=2.5y=56

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  Peter must walk $\begin{array}{r}\frac{5}{6}\end{array}$ of an hour to get to school.
  ::彼得必须步行56小时才能上学

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  Review
  ::回顾

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  For 1-8, write the equation in slope-intercept, point-slope and standard forms.
  ::对于 1-8, 以斜度截面、 点斜面和标准格式书写方程式 。

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  1. The line has a slope of $\begin{array}{r}\frac{2}{3}\end{array}$ and contains the point $\begin{array}{r}\left(\frac{1}{2},1\right)\end{array}$
     ::该线的斜坡为23, 包含点( 12, 1) 。
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  2. The line has a slope of -1 and contains the point $\begin{array}{r}\left(\frac{4}{5},0\right)\end{array}$
     ::线的斜度为-1, 包含点( 45,0)
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  3. The line has a slope of 2 and contains the point $\begin{array}{r}\left(\frac{1}{3},10\right)\end{array}$
     ::该线的斜坡为2, 包含点( 13, 10) 。
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  4. The line contains points (2, 6) and (5, 0).
     ::该行包含点数(2,6)和点数(5,0)。
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  5. The line contains points (5, -2) and (8, 4).
     ::该行包含点数(5,2)和点数(8,4)。
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  6. The line contains points (-2, -3) and (-5, 1).
     ::该行包含点数(2、2、3和5、1)。
  7. 
  8. 

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  For 9-10, solve the problem.
  ::9 -10,解决问题。

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  9. Andrew has two part time jobs. One pays $6 per hour and the other pays $10 per hour. He wants to make $366 per week. Write an equation in standard form that describes this situation. If he is only allowed to work 15 hours per week at the $10 per hour job, how many hours does he need to work per week in his $6 per hour job in order to achieve his goal?
     ::安德鲁有两个兼职工作。 一个每小时付6美元,另一个每小时付10美元。 他想每周挣366美元。 以标准格式写一个方程式描述这种情况。 如果只允许他每周工作15小时,每小时工作10美元,那么他每小时工作6美元需要多少小时才能达到目标?
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  10. Anne invests money in two accounts. One account returns 5% annual interest and the other returns 7% annual interest. In order not to incur a tax penalty, she can make no more than $400 in interest per year. Write an equation in standard form that describes how much she should invest to earn the maximum interest without penalty. If she invests $5000 in the 5% interest account, how much money can she invest in the other account?
      ::Anne在两个账户中投资。 一个账户回报5%的年度利息,另一个账户回报7%的年度利息。 为了不产生税收处罚,她每年的利息不得超过400美元。 以标准格式写一个方程式,说明她应该投资多少来赚取最高利息而不用罚款。 如果她在5%的利息账户中投资5000美元,她可以投资多少钱去另一个账户?

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。