Answer:
=53*(54 ×^9 y^12 z^15)^1/2
=53*y^6 *z^(15/2) * x^( 9/2)* (54)^1/2
Answer:(gf)(3)=g(3)f(3) g(a)=3a+2 or g(3)=3(3)+2=9+2 =11 f(a)=2a−4 f(3)=2(3)−4=6−4=2 g(3)f(3)=112. Answer
Step-by-step explanation:
Answer:
A, because -3 is a negative integer it is to the left of zero, 7 is a positive integer is it to the right of zero.
Hope this helps <3
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
1.25
the exponent is to the 6th because there are six zeros