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Usimov [2.4K]
2 years ago
5

Can u help me pleeeeeeeeeeeease

Mathematics
1 answer:
Ratling [72]2 years ago
8 0
42 for the first one and 105 for the second one
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Suppose a fair 6-sided die is tossed five times. What is the probability that we see at least one three in the five tosses?
Marta_Voda [28]

Answer:

C. 0.1667

Step-by-step explanation:

1/6 +0/6 + -1/6

The closest answer is C. 0.1667

Based on 2nd throw of dice shows zero then  0.1667 of the answer is shown on the 3rd throw in decimal form of -1/6

3 0
2 years ago
11000 is compounded semiannually at a rate of 12% for 21 years. Find the total amount in the compound interest account
marishachu [46]

\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 &\$11000\\
r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus twice}
\end{array}\dotfill &2\\
t=years\dotfill &21
\end{cases}


\bf A=11000\left(1+\frac{0.12}{2}\right)^{2\cdot 21}\implies A=11000(1.06)^{42}\implies A\approx 127127.359416

5 0
3 years ago
The game of Connex contains one 4-unit piece, two identical 3-unit pieces, three identical 2-unit pieces and four identical 1-un
Aleks [24]

Answer: 277 ways

Step-by-step explanation:

Let’s start bycreating 10-unit pieces using the 4-unit piece.

The arrangements are:

1). 4-3-3 (3 permutations)

2). 4-3-2-1 = 4! = 24 permutations.

3). 4-3-1-1-1 (5*[4!/(3!1!)]

= 5*4

= 20permutations

4). 4-2-2-2 (4 permutations)

5). 4-2-2-1-1 (5 *[4!/(2!2!)]

= 5*6

= 30 permutations

6). 4-2-1-1-1-1(6*[5!/(4!1!)]

= 6*5

= 30 permutations

Let’s-consider the arrangements using one or more3-unit pieces and no 4-unit piece:

7). 3-3-2-2 (4!/(2!2!)

8). 3-3-2-1-1 (5*(4!/(2!2!)

= 5*6

= 30 permutations.

9). 3-3-1-1-1-1 (6!/(4!2!) = 0

10). 3-2-2-2-1 (5*4!/(3!1!)

= 5*4

= 20permutations

11). 3-2-2-1-1-1 (6*5!/(3!2!)

= 6*10

= 60 permutations

Finally we would look at the arrangements using only 1-and 2-unit pieces:

12). 2-2-2-1-1-1-1 (7!/(4!3!)

= 35 permutations

add them all up:

(3 + 24 + 20 + 4) + (30 + 30 + 6 + 30) + (15 + 20 + 60 + 35)

=51 + 96 + 130

= 277ways

5 0
2 years ago
Samia created the following tables of values for a linear system. She concluded that there is no
DaniilM [7]

Answer:

Step-by-step explanation:

3 0
3 years ago
Timed please help added picture​
DaniilM [7]

Answer:

\sqrt[4]{2}^{3}

Step-by-step explanation:

7 0
3 years ago
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