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Oxana [17]
3 years ago
8

Dale keeps all of his pairs of shoes in his closet. select the number of shoes that deal could have in his closet. Mark all that

apply. A.3 B.4 C.6 D.7 E.8
Mathematics
2 answers:
bekas [8.4K]3 years ago
6 0
As it says he keeps 'pairs' of shoes so the answer has to be an even 'number' - ie B, C and E are all possible answers.
liberstina [14]3 years ago
6 0
It can Be Either B,C, And E.
A pair of shoe is 2 shoes so this makes the answer even.
In the choice you gave above B,C,and E is even.
So these are the answer
~JZ
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389,422 change all the digits
alexira [117]
389,422 rounded is 400,000.
6 0
3 years ago
bag contains 15 marbles numbered from 1 to 15. Two marbles are selected from the bag at the same time. Let event E = the two num
lbvjy [14]
The numbers DO differ by 10 if they are

- 1 and 11
- 2 and 12
- 3 and 13
- 4 and 14
- 5 and 15

Each outcome technically has two ways of occurring, but since we're taking two marbles at a time, that would be the same as saying, for instance, that drawing 1 and 11 is the same as drawing 11 and 1. We only count such an event once.

So the answer is (a) 5.
7 0
3 years ago
Read 2 more answers
A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest
faltersainse [42]
\bf \qquad \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 &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

\bf \qquad \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 &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396

add both amounts, and that's her investment for the 11 years.
7 0
3 years ago
What's <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B50%20-%201%7D%20" id="TexFormula1" title=" \sqrt{50 - 1} " alt=" \sqrt
nikklg [1K]
Inside the root is 50-1

50-1 = 49

You would remain with root 49 but, if you calculate the square root the answer would be 7 because...

7•7 = 49


^ So, the answer is 7



4 0
3 years ago
Read 2 more answers
Five over seven plus blank over six equals fifty one over forty two
NISA [10]

Answer:

The blank is 3

Step-by-step explanation:

1 - Give the blank a variable

x = our blank

2 - Write it out

\frac{5}{7} + \frac{x}{6} = \frac{51}{42}

3 - Get the denominators equal

\frac{5}{7}(\frac{6}{6}) +\frac{x}{6}(\frac{7}{7}) = \frac{51}{42} \\\frac{30}{42}+\frac{7x}{42} = \frac{51}{42}

4 - Simplify

\frac{30+7x}{42} = \frac{51}{42} \\ \frac{30+7x}{42} = \frac{51}{42}

5 - Solve for the numerators

30+7x = 51

-30         -30

      7x = 21

        x = 3

6 0
3 years ago
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