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andreyandreev [35.5K]
3 years ago
5

Does anyone know the answer?

Mathematics
2 answers:
Helen [10]3 years ago
5 0
The answer would be 60, if you add 13,15, and 12 up you get 40. Those numbers would be considered 2/3 of all the marbles. Divide 40 by 2 to get how many marbles is 1/3. You would get 20 being a third if the marbles. Add that to the original 40 and you’ll get a total of 60.
ddd [48]3 years ago
5 0

Answer: 60

Step-by-step explanation:

Let R be number of red marbels

Let T be Total number of marbels in bag

Now,

T = 13 + 15 + 12 + R

T = 40 + R

Now,

Let P(R) be probability of getting red marble.

P(R) = R / T

1/3 = R / ( 40 + R )

40 + R = 3R

2R = 40

R = 20

So,

T = 40 + R

T = 40 + 20

T = 60

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Six grape fruit cost $5.34. How much does one grape fruit cost????
Firlakuza [10]
In order to figure this out, we can set up a proportion and simplify using cross multiplication.

\frac{units_1}{cost _{1} } =\frac{units_2}{cost _{2} }  \\  \\  \frac{6}{5.34} = \frac{1}{x}  \\  \\ 6*x=5.34*1 \\  \\ 6x=5.34 \\  \\ x=0.89

One grapefruit costs 89 cents.
3 0
3 years ago
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$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F
GuDViN [60]

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}

\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}

\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}

\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762

4 0
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Simplify the inequality​
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Answer:

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