Once the cake is removed from the oven, it has been exposed to the room temperature of 70°F. After 10 minutes, the cake's temperature decreased to 200°F, which is 150°F cooler than it initially was (350°F). As for the what is asked, 90°F is 260°F cooler than its original temperature (350°F).
This problem can be expressed in a ratio:
10mins:150<span>°F = N:260</span><span>°F
where N is how long it will take for</span><span> cake to cool to 90°F
</span>150<span>°FxN=10x260
</span>150N=2600
N=2600/150
N=17.33
N<span>≈20 minutes
Thus, it takes approximately 20 minutes </span><span>for the cake to cool to 90°F</span>
Answer:
Manish has Rs630 while Jhanavi has Rs168.
Step-by-step explanation:
The fraction Manish would have left would be subtracting the fraction on savings as well as that spent on the mall. Which would be;
1 -( 1/2 + 1/4)=1 -3/4 = 1/4
Meaning Manish had 1/4 of his allowance left on him.
Which means 1/4 × Rs. 2520=Rs630
Similarly for Jhanavi, we have :
The fraction left as
1-(1/3 +3/5) = 1 - ( 14/15) = 1/15
Meaning 1/15 of the allowance got remains which is;
1/15 × Rs. 2520= Rs.168
Answer: The change in temperature each hour was 6 degrees celsius.
Step-by-step explanation: 48 divided by 8 = 6
<span>Evan rounded 4.8 down to 4. His estimate of 32 is too low.
I hope this helps, Good luck :)</span>
