1. In an auditorium, there are 22 seats in the first row and 28 seats in the second row. The number of seats in a row continues
2 answers:
Step-by-step explanation:
a) We can find the iterative rule by first making a table of values.
n ->
1 -> 22
2 -> 28
3 -> 34
We see that increases by 6 each time. Hence, the iterative rule should have the term 6n. However, if we put 1 in for n, we get 6. We want 22-6=16 more than this, or
. This would work for any seat row we give.
b) Our iterative rule to get the number of seats is . Since we know that this value is 100, we can put both equal to each other.
Thus, the 14th row has 100 seats.
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I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
