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elixir [45]
3 years ago
6

A stick is broken into two pieces, at a uniformly random break point. Find the CDF and average of the length of the longer piece

.
Mathematics
1 answer:
Oksanka [162]3 years ago
5 0

Answer:

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

Step-by-step explanation:

let assume that stick has length 1.Random variable L that make length of a longer piece and random variable U that mark point .See that L < l means that

U≤ l and 1-U ≤l

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

this means 1-l≤U≤l

so we have

if we have L  [1/2,1]

then apply the formula we have E(L)=3/4

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soldier1979 [14.2K]

Answer:

When both equations have the same slope, but not the same y-intercept, they'll be parallel to each other and no intersections means no solutions. When both equations have different slopes than regardless of the y-intercept they'll intersect for certain, therefore it has exactly one solution.

Step-by-step explanation:

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7 0
2 years ago
A box contains 9 green marbles and 17 white marbles. If the first marble chosen was a white marble, what is the probability of c
pshichka [43]

Answer:

16/25

Step-by-step explanation:

Given

Number  of green marbles = 9

Number of White marbles = 17

Total number of marbles = 26

Probability of picking a white marble = 17/26

When a white marble is picked without replacement,

Probability of picking another white marble after first picking a white marble

= 16/25

= 0.64

This is because the total number of marbles would have reduced from 26 to 25 and the number of white marbles from 17 to 16.

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3 years ago
Reflect the shape over the line shown !
snow_lady [41]
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2 years ago
At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 30 km/h. How fast is
Marrrta [24]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Polly stacked the tins from five boxes of cat food onto an empty shelf in a supermarket. There were 15 tins of cat food in each
Alexandra [31]

Answer:

(a) 3\frac{2}{3} boxes of cat food had been sold.

(b) \frac{11}{3} boxes of cat food had been sold.

Step-by-step explanation:

From the question,

Polly stacked the tins from five boxes of cat food onto an empty shelf and there were 15 tins of cat food in each box. That is,

5 × 15 tins of cat food were stacked onto the empty shelf.

5 × 15 = 75

Hence, 75 tins of cat food were stacked onto the empty shelf.

Also, from the question,

At the end of the week there were 20 tins left on the shelf and the rest of the tins had been sold, that is

75 - 20 tins had been sold

75 - 20 = 55

Hence, 55 tins of cat food had been sold.

To determine the number of boxes of cat food that had been sold,

Since there are 15 tins of cat food per box, then

we will divide 55 tins of cat food that had been sold by 15 tins of cat food per box

55 ÷ 15 = \frac{55}{15}

Hence, \frac{55}{15} boxes of cat food had been sold

(a) As a mixed number

\frac{55}{15}  = 3\frac{2}{3}

Hence, 3\frac{2}{3} boxes of cat food had been sold.

(b) As an improper fraction

\frac{55}{15}  = \frac{11}{3}

Hence, \frac{11}{3} boxes of cat food had been sold.

7 0
3 years ago
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