I am literally so confused please help (i highlighted the triangle by mistake)
2 answers:
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Answer:
E. 0.11
Step-by-step explanation:
We have these following probabilities:
A 10% probability that a person has the flu.
A 90% probability that a person does not have the flu, just a cold.
If a person has the flu, a 99% probability of having a runny nose.
If a person just has a cold, a 90% probability of having a runny nose.
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In this problem, we have that:
What is the probability that a person has the flu, given that she has a runny nose?
P(B) is the probability that a person has the flu. So P(B) = 0.1.
P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.
P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So
What is the probability that this person has the flu?
The correct answer is:
E. 0.11
Answer:
a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx
b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx
c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx
d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)
Step-by-step explanation:
To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.
The chain rule tells you that ...
(f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative
The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.
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a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'
For u = f(x), this becomes ...
sinh(f(x))' = cosh(f(x))·f'(x)
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b) After the same pattern as in (a), ...
cosh(f(x))' = sinh(f(x))·f'(x)
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c) Similarly, ...
tanh(f(x))' = sech(f(x))²·f'(x)
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d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...
sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²
sech(x)' = -sech(x)·tanh(x) . . . . . basic formula
Now, we will use this as above.
sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'
sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)
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Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.
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<em>Additional comment on notation</em>
Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...
f(u)' = f'(u)u'
without getting involved in infinite recursion.
