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

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"

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

tis correct, 7 is 13.46% of 52

Thanks:)

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

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

Oh ok thanks.

yw

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

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

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

yeah...

yeah, your supposed to round to the nearest percent

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

unless you mean, from lbs to kg

no it doesn't say that.

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%

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

oh..

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