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snow_tiger [21]
3 years ago
6

Convert 7896cm to metres.​

Mathematics
2 answers:
Vesna [10]3 years ago
7 0

Answer:

78.96

Step-by-step explanation:

To convert from cm to m you divide the cm by 100.

OleMash [197]3 years ago
6 0

Answer:

78.96

Step-by-step explanation:

You have to count 2 numbers (the 7 and 8) and that’s where your bullet point will go. That’s an easier way to go about it.

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Hlppppppppppppppppppppppppppp
Assoli18 [71]

Answer:

B just did the test

Step-by-step explanation:

8 0
4 years ago
How do I find the equivalent
satela [25.4K]
The equivalent to that is 8.
7 0
3 years ago
How do you do these questions?
Tanzania [10]

Answer:

μ ≈ 2.33

σ ≈ 1.25

Step-by-step explanation:

Each person has equal probability of ⅓.

\left[\begin{array}{cc}X&P(X)\\1&\frac{1}{3}\\2&\frac{1}{3}\\4&\frac{1}{3}\end{array}\right]

The mean is the expected value:

μ = E(X) = ∑ X P(X)

μ = (1) (⅓) + (2) (⅓) + (4) (⅓)

μ = ⁷/₃

The standard deviation is:

σ² = ∑ (X−μ)² P(X)

σ² = (1 − ⁷/₃)² (⅓) + (2 − ⁷/₃)² (⅓) + (4 − ⁷/₃)² (⅓)

σ² = ¹⁴/₉

σ ≈ 1.25

3 0
4 years ago
5. Arrival problems usually follow a Poisson distribution, but in this case the time between arrivals of customers at a bank dur
Finger [1]

Answer:

0.5 or 50%

Step-by-step explanation:

For any given value of 'x' representing the time between arrivals of two customers. If 0 < x <120, then the cumulative distribution function is:

\frac{x-0}{120-0}=\frac{x}{120}

Therefore, the probability that the time between the arrivals of two customers will be more than 60 seconds is determined by:

P(X>60) = 1 -\frac{60}{120}\\P(X>60) = 0.5

The probability is 0.5 or 50%.

5 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
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