5/9(F-32) < 0 or 5/9(F-32)>5

1 answer:

3 0

$\dfrac{5}{9}(F-32) \ \textless \ 0\ \ \ |\cdot\dfrac{9}{5}\\\\F-32 \ \textless \ 0\ \ \ |+32\\\\F \ \textless \ 32\\or\\\dfrac{5}{9}(F-32) \ \textgreater \ 5\ \ \ |\cdot\dfrac{9}{5}\\\\F-32 \ \textgreater \ 9\ \ \ |+32\\\\F \ \textgreater \ 41\\\\Answer:\ F \ \textless \ 32\ or\ F \ \textgreater \ 41\ \to\ F\in(-\infty;\ 32)\ \cup\ (41;\ \infty)$

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Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)