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I am Lyosha [343]
1 year ago
6

568,319,000,000,000,000,000,000,000 in standard form

Mathematics
1 answer:
MatroZZZ [7]1 year ago
3 0

To write in standard form;

568,319,000,000,000,000,000,000,000

Move the decimal point backward till you reach the last number

Multiply by ten raise to the number of times you move the decimal point

That is;

568,319,000,000,000,000,000,000,000 = 5.68319 x 10^26

5.68319\text{  }\times10^{26}

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Step-by-step explanation:

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An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is
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Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...

b) P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

And adding the values we got:

P(X \leq 3) =0.25+0.1875+0.1406=0.578

c) P(X \geq 5) = 1-P(X

And we can find the individual probabilities:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055

P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...

Part b

We want this probability:

P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)

We find the individual probabilities like this:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

And adding the values we got:

P(X \leq 3) =0.25+0.1875+0.1406=0.578

Part c

For this case we want this probability:

P(X \geq 5)

And we can use the complement rule like this:

P(X \geq 5) = 1-P(X

And we can find the individual probabilities:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055

P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316

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