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Leno4ka [110]
3 years ago
5

Simplify 4.2^-3 x 5.1^2 x 4.2^4 x 5.1^3

Mathematics
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

14491.060542

Step-by-step explanation:

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The sequence below shows the number bacteria Arjun observed each hour for a science experiment:
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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]

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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)

a)

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

b)

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

c)

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

d)

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.

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4 years ago
Find the value of 7+c when c=18.
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Step-by-step explanation:

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