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Aneli [31]
1 year ago
13

What negative Z-score (standard score) would the area under the curve be 0.8621z* = _______​

Mathematics
1 answer:
madam [21]1 year ago
7 0
0.8721 = z because its have a asymptote there
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What is the y-value of sin(x) when x= -180°?
Andreyy89

In trigonometry, we start from the point (1,0), so this is the 0-degree point. Then, we start rotating counter clockwise along the unit circle.

Note that rotating 180 or -180 degrees is actually the same, because you travel half the circumference clockwise or counter clockwise. In both cases, you start from the rightmost point of the circle, (1,0), and arrive at the leftmost point, (-1,0).

Now, for every point (x,y) on the unit circle, given by a certain angle \alpha, we have

\cos(\alpha)=x,\quad \sin(\alpha)=y

So, in this case, the angle of -180 is associated with the point (-1,0), and thus we have

\sin(-180)=0

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3 years ago
The data below are the number of absences and the final grades of 9 randomly selected students from a literature class. find the
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3 years ago
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SashulF [63]

Answer:

Use inverse cos to solve for x.

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yuradex [85]

Answer:

42, 44, 45, 46, 48, 49

Step-by-step explanation:

4 0
3 years ago
Help asap plssssssssss
Kryger [21]

Step-by-step explanation:

distance AB is (using Pythagoras with the coordinate differences between both points as sides and the distance as Hypotenuse = baseline of the triangle, the side opposite of the 90 degree angle):

distance² = (5 - -10)² + (7-2)² = 15² + 5² = 225 + 25 = 250

distance = sqrt(250)

to split this distance into a 3/2 ratio, we need actually 3+2=5 equal parts. and then AM gets 3 off these parts, and MB gets 2 of these parts.

so,

distance/5 = sqrt(250)/5 = sqrt(250/25) = sqrt(10)

so,

AM = 3×sqrt(10) = sqrt(90),

and

MB = 2×sqrt(10) = sqrt(40)

now, we need to calculate back using the same Pythagoras approach (calling the coordinates of M xm and ym)

AM² = (xm - -10)² + (ym - 2)² = (xm+10)² + (ym-2)² = 90

90 = xm² + 20xm + 100 + ym² - 4ym + 4

MB² = (5 - xm)² + (7 - ym)² = 40

40 = 25 - 10xm + xm² + 49 - 14ym + ym²

as a first approach we calculate AM² - MB²

90 = xm² + 20xm + ym² - 4ym + 104

- 40 = xm² - 10xm + ym² - 14ym + 74

----------------------------------------------------

50 = 0 + 30xm + 0 + 10ym + 30

20 = 30xm + 10ym

2 = 3xm + ym

ym = 2 - 3xm

this we use now e.g. in the first equation for AM.

90 = xm² + 20xm + (2-3xm)² - 4×(2-3xm) + 104

-14 = xm² + 20xm + 4 - 12xm + 9xm² - 8 + 12xm

-10 = 10xm² + 20xm

-1 = xm² + 2xm

xm² + 2xm + 1 = 0

solving such a squared equation

xm = (-b ± sqrt(b² - 4ac))/(2a)

a = 1

b = 2

c = 1

xm = (-2 ± sqrt(4 - 4))/2 = -1

only one combined solution (as a squared equation usually has 2 solutions).

ym = 2 - 3xm = 2 - 3×-1 = 2 + 3 = 5

so, M = (-1, 5)

8 0
2 years ago
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