Question #47e8d

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Question #47e8d

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Answer 1 · 2018-01-23T22:48:01

28%

Explanation:

If The temperature is 100,
$100$ $100$ decreased by $60 % = 40$ $60%=40$ Degrees
$80 % o f 40 : 32$ $80% of 40: 32$
$40 + 32 = 72$ $40+32=72$
$100 - 72 = 28$ $100-72=28$
$28 %$ $28%$

Answer 2 · 2018-01-23T22:50:36

See a solution process below:

Explanation:

First, let's call the initial temperature: $t$ $t$

If it first decreased by 60% then the new temperature ( $t_{n}$ $t_n$ ) would be:

$t_{n} = t - \frac{60}{100} t$ $t_n = t - 60/100t$

$t_{n} = \frac{100}{100} t - \frac{60}{100} t$ $t_n = 100/100t - 60/100t$

$t_{n} = (\frac{100}{100} - \frac{60}{100}) t$ $t_n = (100/100 - 60/100)t$

$t_{n} = \frac{40}{100} t$ $t_n = 40/100t$

$t_{n} = \frac{4}{10} t$ $t_n = 4/10t$

Then if this new temperature increase by 80% the final temperature ( $t_{f}$ $t_f$ ) would be:

$t_{f} = t_{n} + \frac{80}{100} t_{n}$ $t_f = t_n + 80/100t_n$

However, we can substitute $\frac{4}{10} t$ $4/10t$ for $t_{n}$ $t_n$ giving:

$t_{f} = \frac{4}{10} t + (\frac{80}{100} \times \frac{4}{10} t)$ $t_f = 4/10t + (80/100 xx 4/10t)$

$t_{f} = \frac{4}{10} t + \frac{320}{1000} t$ $t_f = 4/10t + 320/1000t$

$t_{f} = \frac{400}{1000} t + \frac{320}{1000} t$ $t_f = 400/1000t + 320/1000t$

$t_{f} = \frac{720}{1000} t$ $t_f = 720/1000t$

$t_{f} = \frac{72}{100} t$ $t_f = 72/100t$

Therefore, the final temperature is 72% of the original temperature.

Or, in total the temperature decreased by 28%

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Question #47e8d

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