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3 years ago

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A thermometer is taken from a room where the temperature is 23oC to the outdoors, where the temperature is −9oC. After one minut

e the thermometer reads 15oC.
(a) What will the reading on the thermometer be after 4 more minutes?
(b) When will the thermometer read −8oC?

1 answer:

Helga [31]3 years ago

3 0

Answer:

$(a)\ 15\textdegree C\\(b)\ After\ 3\ minutes\ and\ 52.5\ Seconds\ from\ start$

Step-by-step explanation:

Temperature in the room $=23\textdegree C$

Decrease in the temperature(outdoors) $=23-15=8\textdegree C$ in $1\ minute$ .

Rate of decrease $=8\textdegree C/minute$

$Let\ after\ x\ minutes\ temperature\ is\ y\\y=23-8x\ \ \ \ y\geq -9\textdegree C$

(a) After $4\ more\ minutes\ total\ time\ is=1+4=5\ minutes$

$y=23-5\times 8\\y=-17\\$

But it is not possible as outdoor temperature is $-9\textdegree C$ so after $-9\textdegree C$ temperature decrease will not occur.

Hence after $4$ more minutes temperature= $-9\textdegree C$

(b) Let after $t\ minutes$ temperature is $-8\textdegree C$

Here $x=t,\ y=-8\textdegree C$

$-8=23-8x\\8x=23+8\\8x=31\\x=\frac{31}{8}\\x=3.875\ minutes=3\ minutes\ and\ 52.5\ seconds$

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The main thing we have to take into account in this question is that we are about to find the probability of a <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

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With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

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$\\ z = \frac{3.2}{6.69726}$

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$\\ P(z>0.48) = 1 - P(z$

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