Answer: y-int = That means the amount of money in Jana's account before she began mowing lawns.
Step-by-step explanation:
The y-intercept represents the initial money that Jana has in his account. Step-by-step explanation: Given the equation of line is . Also, Jana earns $ per week. And is the number of weeks she has been working. If we plug week in the equation . It will give us the y-intercept. That means the amount of money in Jana's account before she began mowing lawns. The y-intercept represents the initial money that Jana has in his account.
Answer:
1
Step-by-step explanation:
When dividing one fraction by another, the best way to solve it is to multiply by the reciprocal. The reciprocal just flipping the fraction over. So you would leave the 6/7 and the multiply by 14/12( because you are multiplying by the reciprocal). When you do that, you then can just multiply the top and the bottom to form one fraction. The numerator is 6*14 = 84, and the denominator is 7*14 = 84. So you are left with 84/84 which anything over itself is equal to one.
Hope this helps!
Hello from MrBillDoesMath!
Answer:
The fourth choice, b = +\- sqrt( sg + a^2)
Discussion:
s = (b^2 - a^2)/g => multiply both sides by "g"
sg = b^2 - a^2 => add a^2 to both sides
sg + a^2 = b^2 => take the square root of each side
b = +\- sqrt( sg + a^2)
which is the fourth choice.
Thank you,
MrB
Answer:
We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:
Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.
we get :
where C(i) is a constant coefficient obviously between 0 and 1.
All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]
So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.
We already know the sum so it is easy to compute the average :
