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3 years ago

Bob and Laquisha have volunteered to serve on the Junior Prom Committee. The names of twenty volunteers,

Mathematics

2 answers:

marishachu [46]3 years ago

6 0

Answer:

Hey there!

1/19 (1/20)=1/380, so the probability is 1/380

Let me know if this helps :)

Law Incorporation [45]3 years ago

3 0

Answer:

1/380

Step-by-step explanation:

Probability of Bob's name drawn first is:

1/20 as there are 20 people with equal chance

And probability of Laquisha's name drawn second is:

1/19 as there are 19 people left with equal chance

Probability of the two events happening is:

1/20*1/19 = 1/380 product of individual probabilities

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krek1111 [17]

Answer:

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Step-by-step explanation:

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pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

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3 years ago

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stich3 [128]

Answer:

Step-by-step explanation:

So, according to the law of Sines, we have

$\frac{Sin(A)}{a} = \frac{Sin(b)}{b}$

In our case, we have S and R, so...

$\frac{Sin(S)}{s} = \frac{Sin(R)}{r}$

And we want to isolate R...

$r = sin^{-1} (\frac{Sin(S) * r}{s} ) = sin^{-1} (\frac{Sin(52) * 32}{26} ) = 75.90$

the angle of R is then 75.90 degrees.

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