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ss7ja [257]
3 years ago
9

A special deck of cards consists of 5 red cards, 20 blue cards, and 25 green cards. Svetlana selects 1 card from the special dec

k 500 times. How many times can she expect to draw a red card?
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
7 0

Answer:The answer is 50

Step-by-step explanation:

The first step is to add up the numbers in order to get the outcome

5 red + 20 blue +25 green =5+20+25 =50

Step 2

Probability =no of events /no of outcome

We were asked to find the probability of red

Where red represent event

And the summation of red ,green and blue represents outcome .

So probability of red =5/50 = 1/10

Step 3

Recall in the question ,we were told that the card was selected 500 times ,so we multiply the probability of getting red by 500

1/10 *500 =50

Therefore the probability of drawing a red card from the special deck 500 times is 50 .

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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
Help, please I would appreciate it.
Veronika [31]
At firt we should solve what is h(0)
h(0)=0^2+2
h(0)=2
So then you should take the h(0) answer to g(x) and the result is g(2)= -2-5=-7
3 0
3 years ago
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