What do all experiments have in common?

Mathematics

1 answer:

Ksivusya [100]3 years ago

5 0

C. there are at least two different ways in which participants are treated

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Let the random variable X denote the number of network blackouts in a day. The

faltersainse [42]

E(X) = 0(0.7) + 1(0.2) + 2(0.1) = 0.2 + 0.2 = 0.4
The expected daily loss due to blackouts = 0.4 * $500 = $200

Var(X) = 0(0.7 - 0.4)^2 + 1(0.2 - 0.4)^2 + 2(0.1 - 0.4)^2 = 0.04 + 0.18 = 0.22
The expected daily variance due to blackouts = 0.22 * $500 = $110

6 0

3 years ago

Explain how place value could be used to find the difference of 4.23 and 2.75

Usimov [2.4K]

Place value can be used to subtract 4.23 and 2.75 because, in order to subtract, you must line up the whole numbers together, the tenths place together, and the hundredths place together.

Lining them up: Whole= W Tenths place= T Hundredths place= H

W T   H

4 . 2   3
- 2 . 7   5
-------------------
   1 . 4 8

3 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago

How do I find the exact product of 379 & 8?

xxMikexx [17]

Product means the result you get after multiplying. So you have to multiply the two numbers. 379*8 = 3032

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3 years ago

How to do exterior angle

SVETLANKA909090 [29]

The exterior angles of a polygon all add to 360, if you need to find an exterior angle when given the number of sides and that all angles are equal you simply do: 360/ number of sides.
When you need to find the exterior angle when given an interior angle, you do: 180- interior angle.

Hope this helps :)

5 0

3 years ago