a) We know that the probability Jane will win is 0.2, and draws is 0.3, which leaves the probability of her losing to be 0.5 (1 - 0.2 - 0.3 = 0.5).
I'll begin by filling in for the first game:
win = 0.2, draw = 0.3, lose = 0.5
Next, we'll fill in for if she wins, draws, or loses the second game. The probabilities would be the same as the first game for the second game.
Win (0.2): win = 0.2, draw = 0.3, lose = 0.5
Draw (0.3): win = 0.2, draw = 0.3, lose = 0.5
Lose (0.5): win = 0.2, draw = 0.3, lose = 0.5
b) To find the probability that Jane will win both games, we need to multiply the probability of Jane winning the first game by the probability of her winning the second game.
0.2 x 0.2 = 0.04
Hope this helps! :)
Answer: 32
Step-by-step explanation:
Given: The energy needed for 1 hour of running for an adult ![=4.5\times10^6\ joules.](https://tex.z-dn.net/?f=%3D4.5%5Ctimes10%5E6%5C%20joules.)
Therefore. the energy needed for seven hours of running=![=7\times4.5\times10^6\ joules.=31.5\times10^6\ joules](https://tex.z-dn.net/?f=%3D7%5Ctimes4.5%5Ctimes10%5E6%5C%20joules.%3D31.5%5Ctimes10%5E6%5C%20joules)
Since, he energy released by metabolism of 1 average candy bar ![=1\times10^6\ joules.](https://tex.z-dn.net/?f=%3D1%5Ctimes10%5E6%5C%20joules.)
Therefore, the number of candy bars an adult need to eat to supply the energy needed for seven hours of running.=![\frac{31.5\times10^6}{1\times10^6}=31.5\approx32](https://tex.z-dn.net/?f=%5Cfrac%7B31.5%5Ctimes10%5E6%7D%7B1%5Ctimes10%5E6%7D%3D31.5%5Capprox32)
Hence, the adult need to eat 32 candies to supply the energy needed for seven hours of running.
Answer:
Option C. x=7
Step-by-step explanation:
we know that
If PQ is parallel to BC
then
triangles APQ and ABC are similar
Remember that
If two figures are similar them the ratio of its corresponding sides is proportional
so
![\frac{AP}{AB}=\frac{AQ}{AC}](https://tex.z-dn.net/?f=%5Cfrac%7BAP%7D%7BAB%7D%3D%5Cfrac%7BAQ%7D%7BAC%7D)
substitute the values
![\frac{x}{x+7+x}=\frac{x-3}{x-3+x+1}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%2B7%2Bx%7D%3D%5Cfrac%7Bx-3%7D%7Bx-3%2Bx%2B1%7D)
Solve for x
![\frac{x}{2x+7}=\frac{x-3}{2x-2}\\ \\x(2x-2)=(2x+7)(x-3)\\ \\2x^{2}-2x=2x^{2}-6x+7x-21\\ \\3x=21\\ \\x=7\ units](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2x%2B7%7D%3D%5Cfrac%7Bx-3%7D%7B2x-2%7D%5C%5C%20%5C%5Cx%282x-2%29%3D%282x%2B7%29%28x-3%29%5C%5C%20%5C%5C2x%5E%7B2%7D-2x%3D2x%5E%7B2%7D-6x%2B7x-21%5C%5C%20%5C%5C3x%3D21%5C%5C%20%5C%5Cx%3D7%5C%20units)
9514 1404 393
Answer:
see below
Step-by-step explanation:
When a figure is rotated about a center point, the central angle formed by any point on the figure and the corresponding image point will be the rotation angle. Every image point is the same distance from the center of rotation that the pre-image point was. No lengths or angles are changed: rotation is a "rigid motion", so the rotated figure is congruent with the original.
__
-Angle AQA' is 60 degrees.
-Triangles ABC and A'B'C' are congruent.
-Angle ABC is congruent to angle A'B'C'. (<em>this is one of the angles of the congruent triangles</em>)
-Segment BC is congruent to segment B'C'
-Segment AQ is congruent to segment A'Q.