To obtain the probability of obtaining heads at least 3 times out of 5 times we recall that in her simulation <span>she
assigned odd digits to represent heads and even digits to represent
tails. Thus, we count the simulated numbers to see in how many numbers
do we have 3 or more odd digits.
Below, the simulated numbers with 3 or more odd digits are bolded.
</span> <span>32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 <span>65311
</span>
We have 6 simulated numbers having 3 or more odd digits.
Therefore, </span><span>P("heads" at least 3 out of 5 times) = 6 / 16 = 3 / 8.</span>
Answer:
F(x)= -19
Step-by-step explanation:
Just plug in ‘-3’ for every ‘x’ in the equation.
4(-3)-7
-12-7
F(x)= -19
<span>P - 3 1/6 = -2 1/2
3 1/6- 21/2
p=2/3 </span>
Answer:
A) -20
Step-by-step explanation:
distribute the 4y and the -4 to the parentheses
16y^2 -16y+16y-20-16y^2
combine like terms
16y^2-16y^2=0
-16y+16y=0
-20
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