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zheka24 [161]
3 years 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]3 years 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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Whats the perimeter and area of this shape?
stich3 [128]

Answer:

<u>60+14.35=88.27</u>

Step-by-step explanation:

5x12 for the rectangle on the right, then for the circle, using the formula A=πr², I get 28.27, which you divide by 2 because you only need half of the circle, and you get 88.27. I calculated the radius by subtracting 12-6 and then dividing that in half because the radius is half of the distance to the other end of the circle. Then you get the radius of 3, then plug it into the formula to get π(3)², which is just 3.14(9), and you get 28.27.

8 0
3 years ago
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PLEASE HURRY PLEASE PLEASE PLEASE
hram777 [196]

Answer:

b

Step-by-step explanation:

make a formula, its the easiest way to solve problems like these also its very efficent

3 0
3 years ago
How many trucks carried only late variety?
murzikaleks [220]

9514 1404 393

Answer:

  • late only: 15
  • extra-late only: 24
  • one type: 43
  • total trucks: 105

Step-by-step explanation:

It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.

For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.

Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.

__

a) 15 trucks carried only late

b) 24 trucks carried only extra late

c) 4+15+24 = 43 trucks carried only one type

d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out

8 0
3 years ago
A state park charges an entrance fee based on the number of people in a vehicle. A car containing 2 people is charged $14, a car
Artyom0805 [142]

Answer:

Therefore $98 is be charged a bus containing 30 people.

Step-by-step explanation:

Given that,

A state park charges an entrance fee based on the number of people in vehicle.

Let the entry fee for the vehicle be E and entry fee for each person be x.

Then

C= E+(P×x)

C= Total charge in $

E= entry fee for a vehicle

P=No. of person

x= Entry charge per person.

Given A car containing 2 people charged $14

C=$14, P=2

∴14= E+(2× x)

⇒E+2x=14.....(1)

Again A car containing 4 people charged $20

C=$20, P=4

∴20= E+(4× x)

⇒E+4x=20.....(2)

Subtract (1) from (2), we get

E+4x-(E+2x)= 20-14

⇒E+4x-E-2x=6

⇒2x=6

⇒x=3

Putting the value of x in equation (1)

E+(2×3)=14

⇒E=14-6

⇒E=8

Therefore E=$8 and x=$3

Next we check whether our assumption is correct or wrong. Putting the value of E and x for third case

Here P= 8

Therefore C= E+(P×x)

                    = 8+(8× 3)

                    =8+24

                   =$32

Therefore our assumption is correct.

Now C=? , P= 30

The charged for the 30 people is

C= $[8+(30×3)]

  =$[8+90]

 =$98

Therefore $98 is be charged a bus containing 30 people.

4 0
3 years ago
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atroni [7]
I think it's a  although I can't really tell.
7 0
3 years ago
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