3 years ago

The horizontal value at which the highest point of a variable's normal curve occurs is also the ____ of the variable.

1 answer:

4 0

The answer to the question mentioned above is "the mean" such as the whole sentence will become "The horizontal value at which the highest point of a variable's normal curve is also the MEAN of the variable". This is the answer.

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Solve<br> x4 − 8x2 + 16 = 0

Leona [35]

Answer:

x=−2

Step-by-step explanation:

7 0

4 years ago

please help it's due today, And please if you don't know the answers don't put any links or apps. And if you do know the answer

liraira [26]

Answer:

1) Yes

2) No

3) Yes

4) No

5) No

6) Yes

7) Yes

8) No

Step-by-step explanation:

Each equation must be increasing or decreasing by a constant amount in order for it to be linear. Graphing these equations, you can see that the linear equations will be a straight line.

5 0

3 years ago

Which statement about graph and horizontal line is true?

Likurg_2 [28]

Answer:

Step-by-step explanation:

hi! a horizontal line's slope is always 0.

hope this helps:)

6 0

3 years ago

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8.546 rounded to the nearest hundredths

djverab [1.8K]

Answer:

the answer is 8500 as 4 is not greater than or equal to 5

6 0

2 years ago

The height h (in feet) of an object t seconds after it is dropped can be modeled by the quadratic equation h = -16t2 + h0, where

lawyer [7]

$\bf h=-16t^2+255\implies h=-16t^2+0t+255
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}\\\\
\begin{array}{lcccl}
h=& -16 t^2& +0 t& +255\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
h= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
h=\cfrac{-0\pm\sqrt{0^2-4(-16)(255)}}{2(-16)}\implies h=\cfrac{\pm\sqrt{16320}}{-32}
\\\\\\
h=\cfrac{\pm\sqrt{8^2\cdot 255}}{-32}\implies h=\cfrac{\pm 8\sqrt{255}}{-32}\implies h=\cfrac{\mp \sqrt{255}}{4}
\\\\\\
h\approx \mp 3.992179856$

5 0

3 years ago

Read 2 more answers