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Romashka-Z-Leto [24]
3 years ago
15

Put these numbers in order from least to greatest 12%, -1/2, 0, -0.22, 0.56

Mathematics
1 answer:
uysha [10]3 years ago
8 0

Answer: -1/2, -0.22, 0, 12%, 0.56

Step-by-step explanation:

-1/2 is equal to -0.50, making it the number with the least value.

-0.22 is closer to 0 than -0.50, meaning it is greater then -0.50 and less than 0.

0 is between the negative and positive numbers, giving it the spot that it has.

12% is equivalent to 0.12, meaning it is more than 0, and less than 0.56, which is the greatest number.

0.56 has more value than any other number in the problem, meaning it goes last in the order.

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It is predicted that by 2050 there will be 10^10 people living on earth. approximatrly will be 10^12 trees. how many trees for e
Nastasia [14]

Interesting question. Does that mean near the end of humanity?

Solution 1:

10^10 = 1,000,000,000

10^12 = 100,000,000,000

100,000,000,000/1,000,000,000 = 100 trees per person

Solution 2:

10^12 / 10^10  ---  10^10 / 10^10

10^2/1 = 10^2 = 100 trees per person

8 0
3 years ago
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Basile [38]
Find the mean of the first four test scores first.
(Add all numbers in set and divide by how many numbers there are)
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Then, find the mean with the 85 added as the fifth term.
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4 0
3 years ago
Read 2 more answers
N Over 33 = 4 Over 11
raketka [301]

Answer:

d. n = 12

Step-by-step explanation:

n/33 = 4/11

cross-multiply:

11n = 132

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5 0
3 years ago
You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp
Katena32 [7]
Let's move like a crab, backwards some.

after 2 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2

after 3 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3

is that enough to pay the $1100?


now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000   <---- standard form.
8 0
3 years ago
Find the value of the expression. 7! ÷ 3! 5,040 5,034 840 24
koban [17]
First, let's review what the "factorial operator"  !   means.

5! = 5*4*3*2*1.

3! = 3*2*1

7! = 7*6*5*4*3*2*1

Then:  

 7!         7*6*5*4*3*2*1       7*6*5*4         5040
----- = ----------------------- = ------------- = ----------- = 5040 (answer) 
 3!              3*2*1                     1                  1


3 0
3 years ago
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