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3 years ago

mark's computer weighs 35.769 pounds. what is the weight of his computer rounded to the nearest hundredth

2 answers:

8 0

So hundreth is the 2nd number after the decimal point
that is the 6
so to round you take the numer before it so
69
if number is >5 round up

if number is <5 round down

9 is >5 so round up
7

35.77 ppounds

UNO [17]3 years ago

5 0

If you round to the nearest hundredth, Marks computer will weigh 35.77 pounds

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the directions on a bottle of vinegar say "mix 1 cup of vinegar with 1 gallon of water to make a cleaning solution". the ratio o

Hoochie [10]

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3 years ago

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Serga [27]

The range of the function is C.

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$7000 is compounded semiannually at a rate of 11% for 21years Find the total amount and round to the nearest cent

photoshop1234 [79]

$\text{Given that }\$7000 \text{ is compounded semiannualy at a rate of }11\% \text{ for 21 years}\\
\\
\text{we know that the amount after t year when compounded is given by}\\
\\
A=P\left ( 1+\frac{r}{n} \right )^{nt}\\
\\
\text{here P is the principal amoount, so }P=7000,\\
\\
\text{r is the interest rate, }r=11\%=0.11\\
\\
\text{n is the number of times in a year, here semiannulay, so }n=2,\\
\\
\text{and t is the time, so }t=21\\
\\
\text{so the amount after 21 years is}$

$A=7000\left ( 1+\frac{0.11}{2} \right )^{2(21)}\\
\\
\Rightarrow A=7000\left ( 1+0.055 \right )^{42}\\
\\
\Rightarrow A=7000\left ( 1.055 \right )^{42}\\
\\
\Rightarrow A\approx 66328.68$

So the amount after 21 years is: $66328.68

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3 years ago

What is the mass (in grams) of 8.11×1026 CO2 molecules?

Kobotan [32]

Answer:

It is 59,275.74 grams

Step-by-step explanation:

• Molecular mass of carbon dioxide

$= (12 \times 1) + (16 \times 2) \\ = 12 + 32 \\ = 44 \: g$

[ molar masses: C = 12 g, O = 16 g ]

• From avogadro's number, it says 1 mole contains 6.02 × 10^23 molecules or atoms.

But 1 mole is equivalent to its molar mass in terms of weight, therefore;

${ \sf{6.02 \times {10}^{23} \: molecules = 44 \: g }}\\ { \sf{8.11 \times {10}^{26} \: molecules =( \frac{8.11 \times {10}^{26} }{6.02 \times {10}^{23} } \times 44) \: g}} \\ \\ = { \boxed{ \boxed{ \bf{59275.7 \: \: g}}}}$

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3 years ago