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Alenkasestr [34]
2 years ago
12

How can I solve this?

Mathematics
2 answers:
kogti [31]2 years ago
7 0

Answer ok so lets say every 1/3 is one and a  half one-acre lawn, of course 2/3 would make 3 lawns, so maybe a full tank can cut 4 and a half lawns

Step-by-step explanation:

sorry if im wrong

Zanzabum2 years ago
3 0
Bthe answer is 56 ! Because it is I’m sorry
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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
Plz help what’s the answer
Maslowich

Answer:

a. 192 b.60 c. 184, 164, 350

Step-by-step explanation:

a.Total of males=88+104

b.Female, Anaerobic=156-96

c. aerobic= 88+96=184

anaerobic=104+60

total=192+158

8 0
3 years ago
The 20% tip on<br> a $42 meal
blondinia [14]

Answer:

$8.4 hskshskshsjsjjsj

3 0
3 years ago
Which point is on the graph of f(x) = 2*5^x
vodomira [7]

Answer:

Option B (1,10)

Step-by-step explanation:

we have

f(x)=2(5^x)

we know that

If a ordered pair is on the graph of f(x) then the ordered pair must satisfy the function f(x)

<u><em>Verify each case</em></u>

case A) (0,0)

For x=0

f(x)=2(5^0)\\f(x)=2(1)=2

Compare the value of f(x) with the y-coordinate of the ordered pair

2\neq 0

therefore

The ordered pair is not on the graph of f(x)

case B) (1,10)

For x=1

f(x)=2(5^1)\\f(x)=2(5)=10    

Compare the value of f(x) with the y-coordinate of the ordered pair

10=10

therefore

The ordered pair is on the graph of f(x)

case C) (0,10)

For x=0

f(x)=2(5^0)\\f(x)=2(1)=2

Compare the value of f(x) with the y-coordinate of the ordered pair

2\neq 10

therefore

The ordered pair is not on the graph of f(x)

case D) (10,1)

For x=10

f(x)=2(5^{10})\\f(x)=2(9,765,625)=19,531,250

Compare the value of f(x) with the y-coordinate of the ordered pair

19,531,250\neq 1

therefore

The ordered pair is not on the graph of f(x)

7 0
3 years ago
Jason is 22 years older than his sister joan. in 9 years, he will be twice as old as joan. how old is each of them now?
Liono4ka [1.6K]
Jason will be 36 and joan will be 27 years old.
3 0
3 years ago
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