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zheka24 [161]
1 year ago
8

The probability that a random smoker will develop a sever lung condition during his or her lifetime is 0.3. we will choose a ran

dom sample of 120 smokers and let x be the number of these smokers that will develop a severe lung condition.
a.in our sample, find the mean and standard deviation of the number of smokers that will develop a sever lung condition. give your answers to two decimal plac
Mathematics
1 answer:
postnew [5]1 year ago
8 0
The mean is 36 and the standard deviation is 5.02.

The mean is given by
μ = np = 120*0.3 = 36.

The standard deviation is given by
σ = √(n*p*(1-p)) = √(120*0.3*0.7) = √25.2 = 5.02.
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the points (0,1), (1,5), (2,25), (3,125) are on the graph of a function which equation represent that function
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f(x) = 5^{x}

note the range is {1, 5, 25, 125 } which can be expressed as

{ 5^{0}, 5^{1}, 5^{2}, 5^{3} }

the exponents being the domain { 0, 1, 2, 3 }

thus f(x) = 5^{x} an exponential function


4 0
1 year ago
The figure below shows a triangular wooden frame ABC. The side AD of the frame has rotted and needs to be replaced: What is the
Svet_ta [14]
The piece of wood is 5.1 inches
4 0
2 years ago
Read 2 more answers
On a certain​ route, an airline carries 7000 passengers per​ month, each paying ​$30. A market survey indicates that for each​ $
KengaRu [80]

Answer:

The ticket price that maximizes revenue is $50.

The maximum monthly revenue is $250,000.

Step-by-step explanation:

We have to write a function that describes the revenue of the airline.

We know one point of this function: when the price is $30, the amount of passengers is 7000.

We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.

Then, we can write the revenue as the multiplication of price and passengers:

R=p\cdot N=(30+x)(7000-x)

where x is the variation in the price of the ticket.

Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.

R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20

We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:

p=30+x=30+20=50

The maximum monthly revenue is:

R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000

3 0
2 years ago
In a film, an actor played an elderly uncle character is criticized for marrying a woman when he is 3 times her age. He wittily
yanalaym [24]

Answer:the uncle is 54 years old.

The woman is 18 years old

Step-by-step explanation:

Let x represent the age of the uncle.

Let y represent the age of the woman that he married.

In the film, an actor played an elderly uncle character is criticized for marrying a woman when he is 3 times her age. This means that

x = 3y

He wittily replies, "Ah, but in 18 years time I shall only be twice her age. It means that

x + 18 = 2(y + 18)

x + 18 = 2y + 36 - - - - - - - - - - 1

Substituting x = 3y into equation 1, it becomes

3y + 18 = 2y + 36

3y - 2y = 36 - 18

y = 18

x = 3y = 3 × 18

x = 54

6 0
1 year ago
Mr. Schwartz builds toy cars. He begins the week with a supply of 85 wheels, and uses 4 wheels for each car he builds. Mr. Schwa
Reptile [31]

When will he have less than 40 wheels left? Well you can make the equation...

85-4x=40

And then solve for x.

85-4x=40

-4x=-45

x=11 1/4 rounded up to 12

answer: He will have less than 40 wheels left after he has built 12 cars.

5 0
2 years ago
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