A) The vertices of the ABCD square are:
(2,2), (1(4), (3,5), and (4,3).
The reflection over the x-axys keeps the same x-coordinate and changes the y-coordinate ot its negative.So, the vertices of the square A'B'C'D' are:(2,-2), (1,-4), (3,-5), and (4,-3).You can
see the figure attached showing the four vertices.
<span>B. Is AB congruent to A'B'? Explain
Answer: Yes, they are congruent. You can affirm that because
reflections are transformations that do not change either sizes or angles, but keep them, so the two squares are congruent.
C. Is the area of ABCD ≈ area of A'B'C'D'? Explain.
Answer:Yes, the two areas are equal since you have shown that the two figures are congruent.
</span>
Answer:
Could you please supply a dot plot?
Step-by-step explanation:
Sorry if this is trouble
Answer:
B
Step-by-step explanation:
There are 7 keys in each octave (C,D,E,F,G,A and B) so it would make sense to divide 52 to 7.
52/7=7 octaves and 3 extra kets
Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.
If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:
E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]
Or, if you're given the expectation and variance of <em>X</em>, you have
Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²
→ E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]
Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.