Q: Find the critical value for c0.99 and n10?

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it has adkdgbsgbsd'g SD

Two reasons: They're often quicker to use and they make for a great filing system. 1) Speed - Suppose I had a set of objects {n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, n13, n14, n15, n16, n17, n18} that I wanted you to add up, but I wanted you to show your work. Would you rather write this down: n1 + n2 + n3 + n4 + n5 + n6 + n7 + n8 + n9 + n10 + n11 + n12 + n13 + n14 + n15 + n16 + n17 + n18, or an equivalent expression, using indexes, like this: ∑(ni, i, 1, 18)? 2) Filing - Suppose I had a 6 X 6 matrix, or array, and I wanted to talk about one specific element in it. If I hadn't done the proper filing, I would be fumbling for words in an effort to describe where it is. "Go two down from the top and 3 over from the left." I'd rather just have them filed and labeled with indexes, like this n3,4.

int n1; int n2; int n3; int n4; int n5; int n6; int n7; int n8; int n9; int n10; int n11; int n12; int n13; int n14; int n15; int n16; int n17; int n18; int n19; int n20; int n21; int n22; int n23; int n24; int n25; int n26; int n27; int n28; int n29; int n30;

There are 3 odd (1,3,5) and 3 even (2,4,6) sides on a 6-sided die. When rolling a 6-sided die, each face has an equal chance of coming up (assuming a perfect die). Because of this, the chance of rolling an odd number once is 50%. The second roll you again have a 50% chance to roll an odd number, which means that the chance to roll two in a roll is (50% * 50%) 25%. 1/2*1/2 = 1/4 = 25% In other words, the chance of rolling an odd number is 1/2 to the Nth power, where N is the number of rolls. Rolling a dice 10 times means a chance of (1/2)n10 or 1/2 *1/2 *1/2 *1/2 *1/2 *1/2 *1/2 *1/2 *1/2 *1/2 = 1/1024 = 0,097%

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