How can you write 1,360/200 as a repeating decimal?

1 answer:

6 0

1360/200

13.6/2

6.8

It is a finite decimal, but if you want repeating, there are zeroes that follow, like so:

6.8000000000000 . . . (and on to infinity)

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Software filters rely heavily on ""blacklists"" (lists of known ""phishing"" URLs) to detect fraudulent e-mails. But such filter

crimeas [40]

Answer:

a) $X \sim Binom(n=16, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

b) $P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Part a

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=16, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by this formula:

$E(X) = np=16*0.2=3.2$

Part b

For this case we want this probability:

$P(X=0)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And using this function we got:

$P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815$

6 0

3 years ago

(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28

Veronika [31]

$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}$