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PilotLPTM [1.2K]
3 years ago
6

Gloria has 2 5/8 ounces of perfume, If she uses one third of it, how much will she have left?

Mathematics
1 answer:
Marina CMI [18]3 years ago
8 0
Gloria will have 2 7/24 ounces (oz.) of perfume left. Hope it helps :)

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7^16/7^12=7^-18/? what is the missing denominator
creativ13 [48]
\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\
-------------------------------\\\\

\bf \cfrac{7^{16}}{7^{12}}=\cfrac{7^{-18}}{x}\implies x\cdot 7^{16}=7^{12}\cdot 7^{-18}\implies x\cdot 7^{16}=7^{12-18}
\\\\\\
x\cdot 7^{16}=7^{-6}\implies x=\cfrac{7^{-6}}{7^{16}}\implies x=\cfrac{7^{-6}\cdot 7^{-16}}{1}\implies x=7^{-6-16}
\\\\\\
\boxed{x=7^{-22}}\implies x=\cfrac{1}{7^{22}}
4 0
3 years ago
32 x 12.71 pls help me
faust18 [17]

Answer:

406.72

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Suppose that taxis in New York are driven an average of 60,000 miles per year with a standard deviation of 11,000 miles. Assume
Gre4nikov [31]
<span>Let n be the number of taxis in NY. The average distance travelled is 60,000 miles, therefore the middle 95% will have the same average as the population, the reason being the mileage is symmetrically distributed about the mean Therefore the total number of miles in one year for the middle 95% is 60,000 * 0.95 * n
</span><span>The range of miles driven by the middle 95% can be found from the empirical rule that says: For a normal distribution, approximately 95% of the data points lie within the range plus and minus 2 standard deviations of the population mean. In this case the range is (60,000-22,000) to (60,000 + 22,000)</span>
3 0
3 years ago
Joanne deposits $4,300 into a one-year CD at a rate of 2.3%, compounded daily.
jenyasd209 [6]

Answer:

5k

Step-by-step explanation:

8 0
3 years ago
Question Details
Sliva [168]

Answer:

Atomic mass unit-

It is represented by ‘amu’. It is equal to the quantity 1/12th mass of an atom of C-12.

Actual mass of one atom of C-12 = 1.9924 x 10-23 gm. = 1.9924 x 10-26 Kg.

1 amu = 1.9924 x 10 -23/12 = 1.66 x 10 -24 gm. = 1.66 x 10 -27 Kg.

Atomic mass of an element = mass of one atom of the element / 1 amu

Actual mass of an element = atomic mass of an element in amu x 1.66 x 10 -24 gm.

For ex-

Actual mass of hydrogen atom = 1.008 x 1.66 x 10 -24 gm. = 1.6736 x 10-24 gm.

Actual mass of oxygen atom = 16.00 x 1.66 x 10 -24 gm. = 2.656 x 10-23 gm.

Average Atomic weight (Mass)-

“The Average Atomic Masses of many elements are determined by multiplying the atomic mass of each isotope by its fractional abundance and adding these values and then dividing it by 100.”

Ex.- Naturally occurring carbon contains three isotopes – C12 (98.892 % abundance), C13 (1.108 % abundance), C14 (2 x 10-10 % abundance).The relative atomic masses of these isotopes are 12.000, 13.00335 and 14.00317 amu respectively.

Average atomic mass = % of I isotope x its atomic mass + % of II isotope x its atomic mass +% of III isotope x its atomic mass/100

= 12 x 98.892 + 13.00335 x 1.108 + 14.00317 x 2 x10-10 /100

Average atomic mass= 12.011 amu

Molar volume –

Molar volume of a substance is volume occupied by 1 mole of that substance .

Molar volume of solid or liquid = molar mass / density

Molar volume of ideal gas at 00C or 273 K and 1 atmosphere pressure is 22.4 litre.

Gram atomic mass(weight) or gram atom –

“Gram atomic mass(weight) of an element is the mass of Avogadro number ( 6.023 x 1023 ) of atoms of that elements in grams.”

Ex –

mass of 1 atom of oxygen = 16 amu = 16 x 1.66 x 10-24 gm

Mass of 6.023 x 1023 atoms of oxygen= 16 x 6.023 x 1023 x 1.66 x 10-24 = 16 gm.

Hence, Gram atomic weight of oxygen = 16 gm.

No. of gram atoms = mass of element in gm / atomic mass of element in gm.

7 0
3 years ago
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