Mathematics
OpenStudy (anonymous):

you estimate that your school is 45ft tall. your school is actually 52ft tall. find the error precent

OpenStudy (jdoe0001):

your school is really 52, you estimated 45, so you were short by 7 units so, what's 7 units in percent of 52? well, $$\large \begin{array}{llll} \#& \%\\ \hline\\ \cfrac{52}{7} = &\cfrac{100}{x} \end{array}$$ solve for "x"

OpenStudy (anonymous):

So is it 13.46? or did i do it wrong?

OpenStudy (jdoe0001):

tis correct, 7 is 13.46% of 52

OpenStudy (anonymous):

Thanks:)

OpenStudy (jdoe0001):

so your estimate was short by 7% of the actual height

OpenStudy (jdoe0001):

you estimate was 7 units or 13.46% of the actual height rather

OpenStudy (anonymous):

Oh ok thanks.

OpenStudy (jdoe0001):

yw

OpenStudy (anonymous):

Cn u help with this question?....A doctor measures a patient's weight to 162lb to the nearest pound. (Find the maximum possible measurements.)

OpenStudy (jdoe0001):

hmmm.. measurements? I assume it means its height... well.... 162lbs would be someone thin about 5 feet tall I guess

OpenStudy (jdoe0001):

if you mean its weight... well, it's given, 162lbs

OpenStudy (anonymous):

yeah...

OpenStudy (anonymous):

yeah, your supposed to round to the nearest percent

OpenStudy (jdoe0001):

percent? but we would need a point of reference, all we know is that the patient is 162lbs lbs = pounds

OpenStudy (jdoe0001):

unless you mean, from lbs to kg

OpenStudy (anonymous):

no it doesn't say that.

OpenStudy (jdoe0001):

then is rather ambiguous..... because we have no point of reference all we can say is that 162lbs is the only amount we have, thus that's 100%

OpenStudy (jdoe0001):

see... dunno --> (Find the maximum possible measurements.) <--- means

OpenStudy (anonymous):

oh..

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