If you flip a fair coin 4 times, what is the probability that you will get exactly 2 tails

Step-by-step explanation:

One way to solve this problem is to figure out how many ways you can get exactly 222 tails, then divide this by the total number of outcomes you could have gotten. Since every outcome has equal probability, this will be the probability that you will get exactly 2 tails.

How many outcomes are there where you get exactly 2 tails? Try thinking of each outcome as a 4-letter word, where the first letter is "H" if the first coin toss was heads and "T" if it was tails, and so on.

So, the number of outcomes with exactly 2 tails is the same as the number of these words which have 2 T's and 2 H's.

How many of these are there? If we treat all the letters as unique, we'll find that there are 4! different arrangements, overcounting 2! times for every time we only switch the T's around, and 2! times for every time we only switch the H's around.

So, there are 4!/2!2! = 6  

4!/2!2!= 6 different outcomes where you get exactly 2 tails.

Altogether, there are 2^4 = 16 total possible outcomes.

So, the probability that you will get exactly 2 tails is 6/16 = 3/8

3/8

Step-by-step explanation:

khan academy

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.