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2 years ago

If p(a) = 0.50, p(b) = 0.60, and p(a intersection

1 answer:

5 0

Use the identity P(A ∪ B) = P(A)+P(B)-P(A ∩ B)

P(A)=0.50
P(B)=0.60
P(A ∪ B) = 0.30
=>
P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
=(0.50+0.60)-0.30
=0.80

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For f(x) = 5x + 1

drek231 [11]

Answer:

A. $f(7) = 5(7) + 1\\\\f(7) = 36$

B. $f^{-1}(x) = \frac{x - 1}{5}$

C. $f^{-1}(7) =\frac{6}{5}$

D. $f(\frac{6}{5}) = 7$

Step-by-step explanation:

A. To solve the first part of the problem we must replace $x = 7$ in the function $f(x) = 5x + 1$

So:

$f(7) = 5(7) + 1\\\\f(7) = 36$

B. In part B we must find the inverse function of $f(x) = 5x + 1$

To find the inverse function do $y = f(x)$

$y = 5x +1$

Now clear the variable x.

$\frac{y - 1}{5} = x$

Replace x with y.

$y = \frac{x - 1}{5}$

Finally

$f^{-1}(x) = \frac{x - 1}{5}$

C. Now we take the inverse function found above and replace $x = 7$

$f^{-1}(7) = \frac{7 - 1}{5}\\\\f(7) = \frac{6}{5}$

D. Now we substitute $x = f(f^{-1}(7))$ in the original function.

$x = f( f^{-1}(7))\\\\f^{-1}(7) = \frac{6}{5}\\\\ x= f(\frac{6}{5} )\\\\f(\frac{6}{5}) = 5(\frac{6}{5}) + 1\\\\f(\frac{6}{5}) = 7$

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