Answer:
The mean number of the students who develop hypertension over a life time is 7.8.
Step-by-step explanation:
For each person, there are only two possible outcomes, either they will develop hypertension, or they will not. The probability of a person developing hypertension is independent of any other person, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
The expected value of the binomial distribution is:
Suppose that the probability that a person will develop hypertension over a life time is 60%.
This means that 
13 graduating students from the same college are selected at random.
This means that 
Find the mean number of the students who develop hypertension over a life time
The mean number of the students who develop hypertension over a life time is 7.8.
Yes....(y-14)(y+14).....................
Answer:
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Answer:
See below
Step-by-step explanation:
y = -6x^2 + 24x-100
solve for 'x'
y + 100 = -6x^2 + 24x = -6(x^2-4x) complete the square for x
(y+100) /(-6 ) = (x-2)^2 -4 add 4 to both sides
(y+100)/(-6) +4 = (x-2)^2 sqrt both sides
sqrt [ ( y+100)/(-6) + 4 ] + 2 = x change x and y
y = ± sqrt [ (x+100)/(-6) + 4 ] + 2 simplify (if needed)
<u>y = ± sqrt( -x/6 -12.666) + 2 </u>
The given equation, x.cosec²x = cot x - d/dx x.cot x, is proved using the product rule of differentials.
In the question, we are asked to show that x.cosec²x = cot x - d/dx x.cot x.
To prove, we go by the right-hand side of the equation:
cot x - d/dx x.cot x.
We solve the differential d/dx using the product rule, according to which, d/dx uv = u. d/dx(v) + v. d/dx(u), where u and v are functions of x.
cot x - {x. d/dx(cot x) + cot x. d/dx(x)}
= cot x - {x. (-cosec²x) + cot x} {Since, d/dx(cot x) = -cosec²x, and d/dx(x) = 1}
= cot x + x. cosec²x - cot x
= x. cosec²x
= The left-hand side of the equation.
Thus, the given equation, x.cosec²x = cot x - d/dx x.cot x, is proved using the product rule of differentials.
Learn more about differentials at
brainly.com/question/14830750
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