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xxMikexx [17]
3 years ago
8

Question 2

Mathematics
1 answer:
Anton [14]3 years ago
6 0

Answer:

CD=(3-\sqrt{3})\ cm

Step-by-step explanation:

step 1

Find the length side AB

In the right triangle ABD

sin(30\°)=\frac{AB}{AD}

AB=sin(30\°)(AD)

we have

AD=2\sqrt{3}\ cm

sin(30\°)=\frac{1}{2}

substitute

AB=\frac{1}{2}(2\sqrt{3})

AB=\sqrt{3}\ cm

step 2

Find the length side BD

In the right triangle ABD

cos(30\°)=\frac{BD}{AD}

BD=cos(30\°)(AD)

we have

AD=2\sqrt{3}\ cm

cos(30\°)=\frac{\sqrt{3}}{2}

substitute

BD=\frac{\sqrt{3}}{2}(2\sqrt{3})

BD=3\ cm

step 3

Find the length side BC

In the right triangle ABC

we know that

BC=AB -----> is an 45°-90°-45° triangle

therefore

BC=\sqrt{3}\ cm

step 4

Find the length side CD

we know that

BD=BC+CD\\CD=BD-BC

substitute the values

CD=(3-\sqrt{3})\ cm

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Step-by-step explanation:

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P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

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C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

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