Answer:
The sample space would be infinite
{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}
1/8
Step-by-step explanation:
The sample space would be infinite
{HH, TTTHH, TTTTHH, TTTTTTTHH.......................}.
This is because the number of times the coin would be flipped is not specified. So, you can keep flipping the coin forever.
If the coin is tossed four times then the sample space would be
{HHHH HTHH THHH HTHT
HHHT HTTH TTHH THTH
HHTT HHTH TTTH THHT
HTTT TTTT TTHT THTT}
HTHH and TTHH are the only two cases where two consecutive tosses will result in two heads
Probability that the coin will be tossed four times is 2/16 = 1/8
<span><span>
The correct answers are:</span><span>
(1) The vertical asymptote is x = 0
(2) The horizontal asymptote is y = 0
</span><span>
Explanation:</span><span>(1) To find the vertical asymptote, put the denominator of the rational function equals to zero.
Rational Function = g(x) = </span></span>

<span>
Denominator = x = 0
Hence the vertical asymptote is x = 0.
(2) To find the horizontal asymptote, check the power of x in numerator against the power of x in denominator as follows:
Given function = g(x) = </span>

<span>
We can write it as:
g(x) = </span>

<span>
If power of x in numerator is less than the power of x in denomenator, then the horizontal asymptote will be y=0.
If power of x in numerator is equal to the power of x in denomenator, then the horizontal asymptote will be y=(co-efficient in numerator)/(co-efficient in denomenator).
If power of x in numerator is greater than the power of x in denomenator, then there will be no horizontal asymptote.
In above case, 0 < 1, therefore, the horizontal asymptote is y = 0
</span>
Answer:
-5(1-I) is the answer
Step-by-step explanation:
(1+2i)(1+3i)
1(1+3i)+2i(1+3i)
1+3i+2i+6i^2
As i^2=-1
1+5i+6(-1)
1-6+5i
-5+5i
Taking -5 as common
-5(1-i)
I hope this will help you :)
Answer:
50
Step-by-step explanation:
Answer:15-A
Step-by-step explanation: