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Goshia [24]
3 years ago
5

Name 2 fractions that are greater than 1/2. Use benchmark fractions to help

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

2/3, 3/4, 5/6, 7/8, 8/9, 9/10, etc.

Step-by-step explanation:

1/2 is smaller than 1/3, which is smaller than 3/4, which is in turn smaller than 5/6, and so on.

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The table below shows four systems of equations
vazorg [7]
First Answer. 6x-2y=16. This added to 4x-5y=2 is equal to 10x-7y=18
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3 years ago
Two wires, a = 90 ft apart, tether a balloon to the ground, as shown. How high is the balloon above the ground? (Round your ans
valentina_108 [34]

Answer:

180

Step-by-step explanation:

7 0
3 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
Evaluate 2/3x -16 when x= 18<br>A.-6<br>B.-4<br>C.2<br>D.4​
const2013 [10]
2/3 of 18 is 12
then 12- 16 is -4
SO B
8 0
3 years ago
on a map, tannersville, chadwick, and bakersville form a triangle. chadwick is 70 miles frkm tannersville and bakersvilke is 90
Tcecarenko [31]
The phrasing of the question implies there are multiple choices. Is this true?
3 0
3 years ago
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