What is the median of Miguel's data
1 answer:
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Remember
(ab)/(cd)=(a/c)(b/d)
and
1kg=1000g=1*10^3g
hmm, I estimate, 4 of them
x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x=
=
=
a million has 9 zeroes
so 24 zeroes is alot that is a <span>septillion, so there are 2.5316 septillion uranium atoms in 1kg
a. 4
b. underestimate
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(ab)/(cd)=(a/c)(b/d)
and
1kg=1000g=1*10^3g
hmm, I estimate, 4 of them
x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x=
a million has 9 zeroes
so 24 zeroes is alot that is a <span>septillion, so there are 2.5316 septillion uranium atoms in 1kg
a. 4
b. underestimate
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