The answer is (<span><span>3.6<span>x^3 </span></span>− <span>3.766667x</span></span>+<span><span>−1/</span><span>6)
Hope this helps</span></span>
(9,-4) you have to plugin 9 to the equation in the form of x. Then you add 5 and that would be your y value
With n tosses, there are a total of 2^n possible outcomes. We need 15 of theses to be heads.
Since the tosses are random, we want to count the ways to have exactly 15 of them being heads. That will be C(n,15) ways to pick which tosses are heads.
The probability will be given by:
C(n,15)/2^n=0.0148
At this point we use various values of n until we get the correct probability.
n=18⇒0.003112793
n=19⇒0.007392883
n=20⇒0.014785767
n=21⇒0.025875092
Answer:
n=20
Answer:
Step-by-step explanation:
So when you're solving for y you want to get y by it self meaning there should be nothing with y, including any coefficients. The first step is to usually move any terms to the other side and then divide by the coefficient of y so y "technically doesn't" have a coefficient, while it does, but that coefficient 1, so it doesn't really mean much.
Subtract x from both sides
Divide both sides by 6
![13 \sqrt[4]{x} -10 \sqrt[4]{2y}](https://tex.z-dn.net/?f=13%20%5Csqrt%5B4%5D%7Bx%7D%20-10%20%5Csqrt%5B4%5D%7B2y%7D%20)