A coin is tossed 3 times. What is the probability that the number of tails obtained will be between 1 and 3 inclusive?
1 answer:
Answer:
Step-by-step explanation:
Given : When a coin is tossed twice , the total outcomes will become
HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT
Total outcomes = 8
Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT
Number of favorable outcomes = 7
Probability that the number of tails obtained will be between 1 and 3 inclusive =
Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive
You might be interested in
Answer:
D. 2019
Step-by-step explanation:
So I started by rewriting the function f(x)
1000 = 10^3 so
f(x)=10^3 times 1.03^x
1.03 = 103/100 so
f(x)= 10^3 times (103/100)^x
to raise a fraction to a power you just raise the numerator and denominator to that power so
f(x)= 10^3 times 103^x/100^x
change it to exponential form with a base of 10
f(x)=10^3 times 103^x/10^2x
then you can reduce it with the 10^3
f(x)=103^x/10^2x-3
so now that thats in more of a standard form you can find the intersection of the two functions
which is at (9.012,1305.244)
x is the number of years after 2010, so they will be equal ~9 years after 2010, which is 2019.
