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Compute P({1, 2, 3, 4, 5}) − P({1, 2, 3, 4})...

Question

Compute P({1, 2, 3, 4, 5}) − P({1, 2, 3, 4})

compute P({1, 2, 3, 4, 5}) − P({1, 2, 3, 4})



Answers

Compute, as indicated. $$-\frac{1}{5}\left(\frac{3}{4}\right)$$

So we want to be beautiful and well with your arm extractions as complex fractions and our subjection as edition. So we get a negative 7/2, plus a 17/4. Well, that's most. Why would this right to over to? So we gotta get a 14/4. 17 is larger, so we get a positive 17/4 minus a fortune over four. So that gives us a 3/4.

The first thing I'll do is decide on my sign. I know this is positive because they're both negative that I don't have to worry about it throughout the problem. Second thing I'll do is change them into improper fractions. So two times three is six plus one gives me seven and on the right. Four times for 16 plus one gives me 17. Now that I've done that, now I can multiply. Um, there's nothing to reduce here. So seven times 17 is 1 19 over eight. I know 14 times a is 1 12 So this is going to be 14. And then 1 19 minus 1 12 gives me seven so 14 and seven ace.

Okay, so we want to compute the bowling. Well, let's know that we have a minus an a plus sign in its A minus five times to Would you? The 10 minus a minus on the plus on here. So that's a minus. Three times forward at the 12. Okay, so let's write this as edition. So we have negative 10 plus a 12. It's logical to a to

Yes, that we have a positive five times in a day or two, so that's going to be a negative 10 minus a negative. Negative. That's a positive. Four times five, which is 20 all over the Get A four minus two. That's in minus six, so we get a negative 30 over a negative six. We have two minus signs, so not turned into a posture sign. And then we had 30 which is a six times a five all over six. Now we can cancel out our six, and we see that we're left with a five.


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