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3 years ago

Help??? 20 Points

Mathematics

1 answer:

Dominik [7]3 years ago

4 0

1. No; it is not written as a number times the power of 10.

2. 1.22×10^9

3. 2.87×10^−2

4. 5000

5. is answered in attached picture.

6. 3.212×10^3

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3 years ago

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I'll give brainliest!!! please help!

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Answer:

Puritans believed that people (children in particular) were inherently bad (sinners) and had to learn to behave. Salvation lay in learning to check one's natural instincts and behave as an adult.

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3 years ago

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3 years ago

Solve the following system of equations graphically. y=3x+4 y=-2x-5

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2 years ago

In one town, the number of burglaries in a week has a poisson distribution with a mean of 1.9. find the probability that in a ra

tester [92]

Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9

We have to find the probability that in a randomly selected week the number of burglaries is at least three.

P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........

= 1 - P(X < 3)

= 1 - [ P(X=2) + P(X=1) + P(X=0)]

The Poisson probability at X=k is given by

P(X=k) = $\frac{e^{-mean} mean^{x}}{x!}$

Using this formula probability of X=2,1,0 with mean = 1.9 is

P(X=2) = $\frac{e^{-1.9} 1.9^{2}}{2!}$

P(X=2) = $\frac{0.1495 * 3.61}{2}$

P(X=2) = 0.2698

P(X=1) = $\frac{e^{-1.9} 1.9^{1}}{1!}$

P(X=1) = $\frac{0.1495 * 1.9}{1}$

P(X=1) = 0.2841

P(X=0) = $\frac{e^{-1.9} 1.9^{0}}{0!}$

P(X=0) = $\frac{0.1495 * 1}{1}$

P(X=0) = 0.1495

The probability that at least three will become

P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]

= 1 - [0.2698 + 0.2841 + 0.1495]

= 1 - 0.7034

P(X ≥ 3 ) = 0.2966

The probability that in a randomly selected week the number of burglaries is at least three is 0.2966

5 0

3 years ago

Help??? *****20 Points*****

Help??? 20 Points