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3 years ago

What is the least common denominator (LCD) for these fractions? 2/7 and 9/10

Mathematics

1 answer:

yanalaym [24]3 years ago

4 0

Answer:

Step-by-step explanation:

7 and 10 = 70, This is the correct answer because 7 times 10 equals 70, and 10 times 7 equals 70 too, so 70 is the LCD

7 and 10 = 1, This is impossible because anything multiplied by 7 naturally does not get 1 as a answer, including 10 too. X

7 and 10 = 10, Only 10 times 1 equals 10, but 7 a natural whole number does not equal 10. X

7 and 10 = 35, Only 7 times 5 equals 35, but 10 times a whole number does not equal 35. X

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3 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$

Substituting,

$E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}$

Factoring out the n and one p from the above expression:

$E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}$

Representing k=x-1 in the above gives us:

$E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}$

This can then be written by the Binomial Formula as:

$E[ X ] = (np) (p +(1 - p))^{n -1 }= np.$

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3 years ago