Analyze Corey’s work to find his error. What is Corey’s mistake? He substituted 80 for x instead of y. He should have added 88.21 to both sides. He should have rounded down to 32.1°F. He should have predicted about 40 swimmers when the temperature is 80°F.

x is the temperature and y is the number of swimmers. If there are 80 swimmers, then y = 80 instead of x = 80

This is how Corey should have solved

y = 1.505*x - 88.21

80 = 1.505*x - 88.21 <--- replace y with 80

1.505*x - 88.21 = 80

1.505*x = 80 + 88.21 <--- add 88.21 to both sides

1.505*x = 168.21

x = 168.21/1.505 <--- divide both sides by 1.505

x = 111.767

x = 111.8

If there are 80 swimmers at the pool, then the possible outside temperature is roughly 111.8 degrees F. This temperature seems awfully large, so it's possible that Corey's regression equation has limited scope; in other words, it isn't too useful if x is outside a given interval

PLEASE SEE THE ATTACHMENT

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}