X: number of 18-20-year-olds that consume alcoholic beverages in a sample of 50.
The proportion of underage people that drinks are known to be p= 0.697
This variable is discrete. This experiment has two possible outcomes success or failure, we will call "success" each time we encounter an underage individual that consumes alcohol and "failure" will be counting an underage that does not consume alcohol. The number of repetitions of the trial is fixed n= 50. All randomly selected underage individuals are independent and the probability of success is constant trough the whole experiment p=0.697.
Then we can say that this variable has a binomial distribution and we will use that distribution to do the calculations.
a. Under a binomial distribution, the expected value is calculated as:
E(X)= n*p= 50*0.697= 34.85.
The variance of a binomial distribution is:
V(X)= n*p*(1-p)= 50*0.697*0.303= 10.55955
And the standard deviation is the square root of the variance:
√V(X)= 3.2495 ≅ 3.25
b. To know how rare the value 45, you have to see how distant it is concerning the expected value. For this you have to subtract the expected value and divide it by the standard deviation:
[X-E(X)]/√V(X)
(45-34.85)/3.25= 3.12
The value X=45 is 3.12 standard deviations above the mean, which means that it would be rare to find 45 people or more than consumed alcohol.
a) density of the object is 3995.01, b) the weight scale reads 22N c) the sum individually will be the same with when added together.
Step-by-step explanation:
The weight of the object in air is 8N,
and weight = Mass * acceleration due to gravity = m * 9.81
8/9.81 = 0.815,
upthrust( force acting on the body from the liquid impeding the immersion) on the body when fully submerged = weight in air - weight in water = 8N - 6N =2N
Upthrust = weight of water displaced = 2N = mass * acceleration
2/9.81 = 0.204kg
density of water(1000kg/m^3) = mass of water / volume of water
volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)
volume of water displaced = volume of the solid
density of solid = mass/ volume = 0.815/0.000204 = 3995.01kg/m^3
b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N
c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.
