15.
已知四面体ABCD中,E,F分别是AC,BD的中点,若AB=4,CD=2,EF⊥AB,则EF与CD所成角的度数为( )
| A. | 90° | B. | 45° | C. | 60° | D. | 30° |
14.阅读如图所示程序框图,若输入的x=3,则输出的y的值为( )

| A. | 24 | B. | 25 | C. | 30 | D. | 40 |
9.将曲线$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1按φ:$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$变换后的曲线的参数方程为(θ为参数)( )
| A. | $\left\{\begin{array}{l}{x=3cosθ}\\{y=2sinθ}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=\sqrt{3}cosθ}\\{y=\sqrt{2}sinθ}\end{array}\right.$ | ||
| C. | $\left\{\begin{array}{l}{x=\frac{1}{3}cosθ}\\{y=\frac{1}{2}sinθ}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=\frac{\sqrt{3}}{3}cosθ}\\{y=\frac{\sqrt{2}}{2}sinθ}\end{array}\right.$ |
8.送快递的人可能在早上6:30-7:30之间把快递送到张老师家里,张老师离开家去工作的时间在早上7:00-8:00之间,则张老师离开家前能得到快递的概率为( )
0 230966 230974 230980 230984 230990 230992 230996 231002 231004 231010 231016 231020 231022 231026 231032 231034 231040 231044 231046 231050 231052 231056 231058 231060 231061 231062 231064 231065 231066 231068 231070 231074 231076 231080 231082 231086 231092 231094 231100 231104 231106 231110 231116 231122 231124 231130 231134 231136 231142 231146 231152 231160 266669
| A. | 12.5% | B. | 50% | C. | 75% | D. | 87.5% |