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gtnhenbr [62]
3 years ago
14

The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the proba

bility of an occurrence is the same in any two-time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3. The probability that there are 8 occurrences in ten minutes is _________.
Mathematics
1 answer:
Olegator [25]3 years ago
3 0

Answer:

The probability that occurrence of 8 in ten minutes is 0.0771.

Step-by-step explanation:

Poisson Distribution:

A discrete random variable X having the enumerable set {0,1,2,.....} as the spectrum, is said to have Poisson distribution with parameter \mu (>0), if the p.m.f is given by

P(X=x)=\frac{e^{-mu}\mu^x}{x!}  for x=0,1,2,...

               = 0,           elsewhere

The mean number of occurrences in ten minutes is 5.3.

Here \mu = 5.3 and x= 8

P(X=8)=\frac{e^{-5.3}(5.3)^8}{8!}

               =0.0771

The probability that occurrence of 8 in ten minutes is 0.0771.

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By paying $50 cash up front and the balance at $35 a
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Answer:

26 weeks i think

Step-by-step explanation:

6 0
3 years ago
How much more did Isabel earn than Sarah?
N76 [4]

where's the storyy sisssssss

8 0
3 years ago
A catering service offers
Fiesta28 [93]

Answer:

There are 52920 ways to do the banquet

Step-by-step explanation:

* Lets revise the combination to solve this problem

- A combination is a collection of the objects where the order

 doesn't matter

- The order is not important in this problem, so we can use the  

  combination nCr to find how many ways can this done

* Lets solve the problem

- A catering service offers 6 appetizers, 7 main courses, 10 desserts.

- A customer is to select 5  appetizers,  4  main courses, 5  desserts

 for a banquet

∵ We want to select  5  appetizers from 6

∴ There are 6C5 ways to chose the appetizers

∵ 6C5 = 6

∴ There are 6 ways to chose the appetizers

∵ We want to select  4  main courses from 7

∴ There are 7C4 ways to chose the main courses

∵ 7C4 = 35

∴ There are 35 ways to chose the main courses

∵ We want to select  5  desserts from 10

∴ There are 10C5 ways to chose the desserts

∵ 10C5 = 252

∴ There are 252 ways to chose the desserts

- To find the number of ways to do the banquet multiply all the

  answers above

∵ The total ways = 6C5 × 7C4 × 10C5

∴ The total ways = 6 × 35 × 252 = 52920

* There are 52920 ways to do the banquet

5 0
3 years ago
Sally is 1 year older than 3 times the age of joe.Chirs is 5 years younger than 6 times the age of joe.If sally and Chris are th
Sav [38]
S = sally C = Chris J = Joe
sally is one year older than 3 times the age of joe, so
s = 3j + 1
chris is 5 years younger than 6 times the age of jo, so
c = 6j - 5
sally and chris are the same age so
s = c so
3j + 1 = 6j - 5
-3j -3j
1 = 3j - 5
+5 +5
6 = 3j or
3j = 6
3j/3 = 6/3
j = 2

check
s = c
3(2) +1 = 6(2) - 5
6 + 1 = 12 - 5
7 = 7 correct✅
7 0
4 years ago
Read 2 more answers
50 Points!
Advocard [28]
Well, 50 points total given, 25 points per user  and 13 bonus for brainliest

anyway


1.
add them equations, the y's will cancel
x+2y=4
<span>3x-2y=4 +</span>
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution


2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
<span>-4x+14y=-6 +</span>
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions



3.
multiply first equation by 2 and add to first
4x+2y=-6
<span>1x-2y=-4 +</span>
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)


4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16


5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'


EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a


40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
<span>-120=-3c-3a +</span>
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold


ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
7 0
4 years ago
Read 2 more answers
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