Answer:
3,340$
Step-by-step explanation:
e(8000)=.0675(8000)+2800
e(8000)=540+2800
e(8000)=3340
In order to solve or know the probability of having 2 girls
and 2 boys, assumed that a girl is as likely as a girl at each birth, pascal’s
triangle will be likely used. And we will be referring to the line 4 of pascal’s
triangle, which was 1 4 6 4 1. Then it
will look like this: 1 = 4 girls; 4 = 3 girls & 1 boy; 6 = 2 girls & 2
boys; 4 = 3 boys & 1 girl; 1 = 4 boys. And now for the solution in order to
get the probability of having 2 girls and 2 boys is to divided into the sum of 1+4+6+4+1.
Answer:
A. Rotation 90º clockwise about the origin B. Rotation 180º about the origin C. Rotation 90º counterclockwise about the origin D. Reflection across y=x
Step-by-step explanation:
because
=7(∛2x) - 6(∛2x) - 6(<span>∛x)
= </span>∛2x - 6<span>∛x
answer
C. </span>∛2x - 6∛x
third choice
<span>So you have composed two functions,
</span><span>h(x)=sin(x) and g(x)=arctan(x)</span>
<span>→f=h∘g</span><span>
meaning
</span><span>f(x)=h(g(x))</span>
<span>g:R→<span>[<span>−1;1</span>]</span></span>
<span>h:R→[−<span>π2</span>;<span>π2</span>]</span><span>
And since
</span><span>[−1;1]∈R→f is defined ∀x∈R</span><span>
And since arctan(x) is strictly increasing and continuous in [-1;1] ,
</span><span>h(g(]−∞;∞[))=h([−1;1])=[arctan(−1);arctan(1)]</span><span>
Meaning
</span><span>f:R→[arctan(−1);arctan(1)]=[−<span>π4</span>;<span>π4</span>]</span><span>
so there's your domain</span>
