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3 years ago

Reciprocal of -2/3^6

Mathematics

1 answer:

Elodia [21]3 years ago

3 0

Answer:

The answer is -729/2

Step-by-step explanation:

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Nitella [24]

Answer:

Step-by-step explanation:

5 0

3 years ago

The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe

marissa [1.9K]

Answer:

a) 0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If $F_X(x)$ is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

$P(X \leq x) = F_X(x)$

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

$P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)$

We are looking for P(X<2)

$P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31$

In this case we want P(X>3)

$P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03$

Now, we are interested in P(X=1)

$P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36$

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

$P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13$

$P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03$

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

8 0

4 years ago

Write an expression for "10 multiplied by u."

Free_Kalibri [48]

Hello from MrBillDoesMath!

Answer:

10u

Discussion:

10 multiplied by u is equivalent to 10u

Thank you,

MrB

3 0

4 years ago

Read 2 more answers

Please help with this. Thanks!

Black_prince [1.1K]

Answer:

1/5k-2/3j and -2/3j+1/5k

Step-by-step explanation:

This is because the sign of both of the terms stay the same and the fractions and variables stay the same for each term as well.

6 0

3 years ago

Lim x-->+ infinity (2^x)/x^10

nadezda [96]

$\displaystyle\lim_{x\to\infty}\frac{2^x}{x^{10}}=\lim_{x\to\infty}\frac{e^{x\ln2}}{x^{10}}=\dfrac\infty\infty$

A few applications of L'Hopital's rule gives a decent idea of how this limit will ultimately behave.

$=\displaystyle\lim_{x\to\infty}\frac{\ln2\,e^{x\ln2}}{10x^9}$
$=\displaystyle\lim_{x\to\infty}\frac{(\ln2)^2\,e^{x\ln2}}{90x^8}$

and so on. Notice that the numerator will consistently behave exponentially, while the denominator will eventually be rendered into a constant. This means the function diverges to $\infty$ as $x\to\infty$ .

6 0

3 years ago