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anonymous
 one year ago
Can you help me with 1.65?
anonymous
 one year ago
Can you help me with 1.65?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@toxicsugar22 do you know?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@aaronq can you help?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1You just have to use the formula written there at the bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i understand that, but which is the T.V and the E.V??? that's what i don't understand @aaronq

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1true value and experimental value. Did they not give you these?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what's in the picture is what we got

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you want the full question?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1really? maybe they want you to assume that the body temp is 37 Celsius

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1but yeah, post the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it said body temp was 38.9 C

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1was that supposed to be the measured temperature?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The medicinal thermometer commonly used in the homes can be read (+or) 0.1F, whereas those in the doctor's office may be accurate to (+or) 0.1C, express the percent error expected from each of these thermometers in measuring a person's body temperature of 38.9C.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1ohh okay. So the true value is 38.9 C and the largest possible error is \(\pm0.1~^oC\) So \(\sf percent~error =\dfrac{trueexperimental}{true}*100\%\) note that the absolute value bars and how it wont matter if you add or subtract 0.1 C from the value. For celsius it's: \(\sf percent~error =\dfrac{38.9^oC39~^oC}{38.9^oC}*100\%\)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1Do the same for fahrenheit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what would the exp. be for F? 102.02 is true. so would exp. be 102.01?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1102.02 F +0.1 F = 102.12 F

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm getting .0980% (sorry it took so long, i got a phone call) @aaronq

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1yep, that seems good, i would round to 0.1 % though
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