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Otrada [13]
3 years ago
15

Let x be the number of successes throughout n independent repetitions of a random experiment having probability of success p=1/4

. determine the smallest value of n so that p(1<=x)>=0.7
Mathematics
1 answer:
Llana [10]3 years ago
6 0

To answer problems like this you have to use binomial:

P (x > 1) = 1 – p (0 < x < 1) > .7 

So:

1 – p (0) – p (1) > .7 

1 – (3/ 4) ^n – (3/ 4) ^n (n – 1 ) (1/ 4) > .7 

Therefore n > 5.185, and the smallest value of n so that we can satisfy the given condition is 6 (rounded up)

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3 years ago
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ALL MY QUESTIONS GETTING IGNORED PLS HELP
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Answer:

x=9

Step-by-step explanation:

a scale drawing means all the sides are multiplied by the same amount. to find this amount, you can divide the larger side by the bigger side by the smaller side.

15/5 = 3

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therefore the scale factor is 3. to get the value of x, multiply the value of the third side by 3.

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2 years ago
Ed earns a $100 commission on each computer he sells plus a base salary of $50,000. Which inequality can be used to find how man
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Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D
viva [34]

Answer:

\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C

Step-by-step explanation:

So we have the indefinite integral:

\int (\ln(x))^2dx

This is the same thing as:

=\int 1\cdot (\ln(x))^2dx

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}

Simplify:

du=\frac{2\ln(x)}{x}dx

And:

dv=(1)dx\\v=x

Therefore:

\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx

The x cancel:

=x\ln(x)^2-\int2\ln(x)dx

Move the 2 to the front:

=x\ln(x)^2-2\int\ln(x)dx

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

=x\ln(x)^2-2\int 1\cdot\ln(x)dx

Let u be ln(x) and let dv be (1)dx. Thus:

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And:

dv=(1)dx\\v=x

So:

=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)

Simplify the integral:

=x\ln(x)^2-2(x\ln(x)-\int (1)dx)

Evaluate:

=x\ln(x)^2-2(x\ln(x)-x)

Now, we just have to simplify :)

Distribute the -2:

=x\ln(x)^2-2x\ln(x)+2x

And if preferred, we can factor out a x:

=x(\ln(x)^2-2\ln(x)+2)

And, of course, don't forget about the constant of integration!

=x(\ln(x)^2-2\ln(x)+2)+C

And we are done :)

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