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.

7 0

3 years ago

Which of the following functions is graphed below.

Valentin [98]

Answer: Option A

$y=\left \{ {{x^2 +2;\ \ x$

Step-by-step explanation:

In the graph we have a piecewise function composed of a parabola and a line.

The parabola has the vertex in the point (0, 2) and cuts the y-axis in y = 2.

The equation of this parabola is $y = x ^ 2 +2$

Then we have an equation line $y = -x + 2
$

Note that the interval in which the parabola is defined is from -∞ to x = 1. Note that the parabola does not include the point x = 1 because it is marked with an empty circle " о ."

(this is $x< 1$ )

Then the equation of the line goes from x = 1 to ∞ . In this case, the line includes x = 1 because the point at the end of the line is represented by a full circle .

(this is $x\geq 1$ )