c. Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $12 and $18

Mathematics

1 answer:

Dmitriy789 [7]4 years ago

3 0

Answer:

0.0066

Step-by-step explanation:

The x has a distribution that is approximately normal. For normal distribution the probability of x will be,

u = 12 and 18

P (12 $\leq$ $x$ $\leq$ 18)

P (- 2.3 $\leq$ $x$ $\leq$ 2.85 )

P (0.9912 - 0.9978)

= 0.0066

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The midpoint of \overline{\text{AB}} AB is M(-3, -3)M(−3,−3). If the coordinates of AA are (-5, -1)(−5,−1), what are the coordin

astraxan [27]

The coordinates of point B are (-1,-5)

Step-by-step explanation:

The formula for mid-point is given by:

$M = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})$

Given

M(x,y) = (-3,-3)

A(x1,y1) = (-5, -1)

We have to find the coordinates of B(x2,y2)

So,

$M(x,y)= (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})\\(-3,-3) = (\frac{-5+x_2}{2} , \frac{-1+y_2}{2})\\putting\ respective\ coordinates\ equal\\-3 = \frac{-5+x_2}{2}\\-3 * 2 = -5+x_2\\-6 = -5+x_2\\-6+5 = x_2\\x_2 = -1\\And\\-3 = \frac{-1+y_2}{2}\\-3*2 = -1+y_2\\-6 = -1+y_2\\-6+1 = y_2\\y_2 = -5$