Answer:
B
Step-by-step explanation:
For a relationship to be a function
Each value of x in the domain can only have 1 unique value of y in the range. That is, one-to-one correspondence.
The only relation which meets this criteria is B
Complete question :
Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.
Answer:
0.10935
0.3718
0.9778
0.606
Step-by-step explanation:
μ = 721 ; σ = 454 ; n = 50
P(x > 800)
Zscore = (x - μ) / σ/sqrt(n)
P(x > 800) = (800 - 721) ÷ 454/sqrt(50)
P(x > 800) = 79 / 64.205295
P(x > 800) = 1.23
P(Z > 1.23) = 0.10935
2.)
Less than 700
P(x < 700) = (700 - 721) ÷ 454/sqrt(50)
P(x < 700) = - 21/ 64.205295
P(x < 700) = - 0.327
P(Z < - 0.327) = 0.3718
Between 700 and 850
P(x < 850) = (850 - 721) ÷ 454/sqrt(50)
P(x < 850) = 129/ 64.205295
P(x < 700) = 2.01
P(Z < 2.01) = 0.9778
P(x < 850) - P(x < 700) =
P(Z < 2.01) - P(Z < - 0.327)
0.9778 - 0.3718
= 0.606
Answer:
The solution to the given system of equations is (-2,
)
Therefore the values of x and y are x=-2 and 
Step-by-step explanation:
Given equations can be written as

Solving equation(2) we get
x=-2
Substitute x=-2 in equation (1) we get




Therefore the values of x and y are x=-2 and 
The solution to the given system of equations is (-2,
)
Answer:
The number of bacteria at initial = 187
Step-by-step explanation:
Given that the population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t.


Integrating both side we get
㏑ N = k t + C ------- (1)
Now given that after 3 hours it is observed that 500 bacteria are present and after 10 hours 5000 bacteria are present.
⇒ ㏑ 500 = 3 k + C -------- (2)
⇒ ㏑ 5000 = 10 k + C ------ (3)
⇒ ㏑ 5000 - ㏑ 500 = 7 k
⇒ ㏑
= 7 k
⇒ ㏑ 10 = 7 k
⇒ k = 0.329
Put this value of k in equation (2),
⇒ ㏑ 500 = 3 × 0.329 + C
⇒ C = 5.23
Put this value of C in equation 1 we get,
⇒ ㏑ N = k t + 5.23
Initially when t = 0 , then
⇒ ㏑ N = 5.23
⇒ N = 187
Thus the number of bacteria at initial = 187