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Alex73 [517]
3 years ago
9

What is 6789 / 33please show work you guys

Mathematics
1 answer:
Norma-Jean [14]3 years ago
4 0
Theres how I did it :)

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If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?
Lostsunrise [7]
Since \mathrm M_Y(t)=e^{6(e^t-1}, we know that Y follows a Poisson distribution with parameter \lambda=6.

Now assuming \mu,\sigma denote the mean and standard deviation of Y, respectively, then we know right away that \mu=6 and \sigma=\sqrt6.

So,

\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028
6 0
3 years ago
In rhombus MNPQ, diagonals MP and NQ intersect at point K. If m_MQP = 56', then which of the
Murrr4er [49]

the answer is 90 degrees

3 0
2 years ago
Angle measure represented by 36.7 rotations counterclockwise
zhannawk [14.2K]

Answer:

  • 13,212° or 73.4π radians

Step-by-step explanation:

Each rotation is 360° or 2π radians. So, 36.7 rotations is ...

  36.7×360° = 13,212°

or

  36.7×2π = 73.4π radians

7 0
3 years ago
Please please please help. thank you.
djverab [1.8K]
Lets get started :)


The First question:
Diameter = 20 ft
Radius = \frac{1}{2} of 20 (diameter) = 20 ft

Area formula of a circle is
A = \pir²
    =\pi(10)²
    =100\pi ft²
    ≈ 314 ft²

The answer will be the second option

The Second question:
radius of circle = 12mm  divided into 20 sectors area

A = \pir²
   = \pi(12)²
   = 144\pi ft²

Divide into 20 equal sector areas = \frac{144 \pi }{20} = 7.2 \pi

≈ 22.6 mm²

Your answer will be the third option

The Third Question:

90°, sector area = 36\pi , Radius = ?

\frac{angle}{360} = \frac{sector}{ \pi (r)^2}
\frac{90}{360} = \frac{36 \pi }{ \pi r^2}
\pi and \pi cancels out

We can now cross multiply
360 × 36 = 90r²
12960 = 90r²

Divide by 90 on either side

\frac{12960}{90} = \frac{90r^2}{90}
144 = r²

Take squareroot 
 \sqrt{144} = x
x = 12 in

Your answer will be the third option




4 0
3 years ago
Evaluate. PICTRE BELOW.
ololo11 [35]

Answer:

d ko makita yung pic sorry

4 0
2 years ago
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