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jolli1 [7]
3 years ago
8

london buys 3 more packs of trading cards today. each pack has 8 cards. write a multiplication sentence to show how many cards l

andon buys today. then find how many cards landon has now.
Mathematics
1 answer:
deff fn [24]3 years ago
8 0
The answer to this question is 24
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A number is no less than 13
Yakvenalex [24]

Answer: yes a number is less than 13

Step-by-step explanation:

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Each investment matures in 3 years. The interest compounds annually.
Sveta_85 [38]

bearing in mind that 4¾ is simply 4.75.

\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}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill

\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}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032

well, the interest for each is simply A - P

695.575 - 600 = 95.575.

862.032 - 750 = 112.032.

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3 years ago
How do I solve quadratic using the quadratic formula
Irina18 [472]

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A box contains 2 blue cards numbered 1 through 2, and 3 green cards numbered 1 through 3. List the sample space of picking a blu
Akimi4 [234]

Answer:

See explanation

Step-by-step explanation:

A box contains 2 blue cards numbered 1 through 2. Let them be named B1 and B2. This box also contains 3 green cards numbered 1 through 3. Let them be named G1, G2 and G3.

The sample space of picking a blue card followed by a green card is

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So, there are 6 different outcomes in this sample space.

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