Calculating P(U-3 < 4.33 sqrtV) Using t Distribution

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t distribution??

Homework Statement


If U and V are independent, U being distributed N(3,16) and V being distributed as chi-square on 9 degrees of freedom, find

P(U-3 < 4.33 sqrtV).


Homework Equations





The Attempt at a Solution


Well I don't even know where to start! Does U-3 suggest the mean is 0? Is the whole distribution of U shifted left by 3? Does that give a t distribution?
I am not asking for the answer but some direction would be appreciated, I just can't seem to understand this at all.
 
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U-3 is distributed N(0,16).
 


Ok, I understand this and also can reason that the probability would be very high, just by thinking of the graphs of the distributions, however I am still at a loss as to how to compute the probability. I use R to do the calculations, is it some kind of t test or do I put in the distributions and their parameters and just use R to calculate the probability, if so... is the R code just the same as in the question, that is, P(U-3 < 4.33 sqrtV)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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