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2 years ago

Solve the quadratic equation by using either a numeric or a graphic approach.

Mathematics

1 answer:

dimulka [17.4K]2 years ago

4 0

Answer:

Step-by-step explanation:

x²-14x-49=0

x²-14x+49=49+49

(x-7)²=98

(x-7)²=49×2

x-7=±7√2

x=7+7√2=7(1+√2)≈7(1+1.414)≈7(2.414)≈16.898≈16.9

x=7-7√2=7(1-√2)≈7(1-1.414)≈-7×0.414≈-2.898≈2.9

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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,

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

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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

231 ( 40 + 15 ) = 231 x 40 + ------------ x -----------<br><br> which property is this

a_sh-v [17]

Answer:

12705

Step-by-step explanation:

The problem could be solved in two different methods:

231×(55)=12705

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2 years ago

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dmitriy555 [2]

Answer:

Step-by-step explanation:

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2 years ago