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never [62]
3 years ago
7

About 2% of the population has a particular genetic mutation. 400 people are randomly selected. Find the mean for the number of

people with the genetic mutation in such groups of 400.
Mathematics
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

200

Step-by-step explanation:

So it'll be 400 ÷ 2 = 200 , and a way to to check your way would be 200 x 2 = 400

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The midpoint of (x1,y1) and (x2,y2) is (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

just average them


(4,1) and (-2,4)
the midpoint is  (\frac{4-2}{2},\frac{1-4}{2})=(\frac{2}{2},\frac{-3}{2})=(1,-1.5)
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Find the lowest common denominator for the following fractions.
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Your answer is 6 because 6 is the smallest number that is divisible by both 2 and 3
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What is the solution to this system of linear equations?
svetoff [14.1K]
It’s B
If y-x=6
Y +X =_10
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7 0
3 years ago
The probability that a civil servant own a car is 1/6,if two civil servants are selected at random.find the probability that
Elodia [21]

Answer:

The data we have is:

the probability that a civil servant own a car is 1/6

Two civil servants are selected at random.

a) The probability that each own a car.

Ok, here we have two events:

Person 1 haves a car.

Person 2 haves a car.

The probability of each of those events is the same, 1/6.

P1 = 1/6

P2 = 1/6

Now, the probability of both events happening at the same time is equal to the product of the individual probabilities.

P = P1*P2 = (1/6)*(1/6) = 1/36.

b) Only one has a car.

Suppose that Person 1 has the car and Person 2 has not a car.

The probability for person 1 is 1/6.

And for person 2, is the negation of having a car:

So if we have prob of having a car = 1/6

Then the probability of not having a car is = 1 - 1/6 = 5/6.

Then we have:

P1 = 1/6

P2 = 5/6.

The joint probability is:

P = P1*P2 = (1/6)*(5/6) = 5/36.

But we also have the case where person 1 does not have a car, and person 2 does have one, then we have a permutation, and the actual probability is two times the obtained above.

P = 2*(5/36) = 10/36

4 0
3 years ago
Data Entry - No Scoring During your procedure to determine the uncalibrated volume near the stopcock of your buret, what was the
DaniilM [7]

Answer:

0.025 grams

Step-by-step explanation:

The water in the stopcock has a volume of 25 mL initially, After that, the whole water was drained out. So we have:

Volume of drained water = (25 mL)(1 x 10⁻⁶ m³/1 mL)

Volume of drained water = 25 x 10⁻⁶ m³

Density of drained water = 1000 kg/m³

So, for the mass of drained water:

Density of drained water = Mass of drained water/Volume of drained water

Mass of drained water = (Density of drained water)(Volume of drained water)

Mass of drained water = (1000 kg/m³)(25 x 10⁻⁶ m³)

<u>Mass of drained water = 0.025 gram</u>

Density

4 0
3 years ago
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