题目


A utility company sells electricity to its customers by units of energy called kilowatt hours (kWh). In order to encourage its customers to use less electricity during periods of peak electrical demand in the summer, the company charges its customers a higher amount per kWh for electricity during peak usage times. The graph shows the amount, in USD per kWh, that the company charges its customers for electricity during both the summer and non-summer months, from the hours of 6:00 a.m. until 6:00 p.m.


Select from each drop-down menu the option that creates the most accurate statement based on the information provided.


During the non-summer months, customers of the utility company will pay {【下拉选项标识】} percent more per kWh used at 8:00 a.m. than they would during the summer months, and would pay {【下拉选项标识】} percent less per kWh at 3:00 p.m. than they would during the summer months.

解析

本题需要根据图表数据,分别计算“非夏季(non-summer)”与“夏季(summer)”在对应时间的电价百分比变化。 #### 1. 计算8:00 a.m.的“多付百分比” - 夏季(蓝色柱)8:00 a.m.电价:\( 0.04 \) USD/kWh - 非夏季(黑色柱)8:00 a.m.电价:\( 0.06 \) USD/kWh 百分比变化公式(“A比B多付的百分比”): \[ \text{多付百分比} = \frac{\text{非夏季电价} - \text{夏季电价}}{\text{夏季电价}} \times 100\% \] 代入数据: \[ \frac{0.06 - 0.04}{0.04} \times 100\% = \frac{0.02}{0.04} \times 100\% = 50\% \] #### 2. 计算3:00 p.m.的“少付百分比” - 夏季(蓝色柱)3:00 p.m.电价:\( 0.12 \) USD/kWh - 非夏季(黑色柱)3:00 p.m.电价:\( 0.06 \) USD/kWh 百分比变化公式(“A比B少付的百分比”): \[ \text{少付百分比} = \frac{\text{夏季电价} - \text{非夏季电价}}{\text{夏季电价}} \times 100\% \] 代入数据: \[ \frac{0.12 - 0.06}{0.12} \times 100\% = \frac{0.06}{0.12} \times 100\% = 50\% \] ### 结论 8:00 a.m.非夏季比夏季多付50%,3:00 p.m.非夏季比夏季少付50%,因此两个空的答案均为50。
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