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Shkiper50 [21]
3 years ago
10

Your answer is incorrect. Suppose that a "code" consists of 6 digits, none of which is repeated. (A digit is one of the 10 numbe

rs 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 .) How many codes are possible?

Mathematics
2 answers:
yarga [219]3 years ago
8 0
Use factorials to solve this problem.

When you are choosing a number of digits from a set without repetition, you will use the following formula:

\frac{n!}{(n-r)!}

n represents the total amount of items in the set, and r represents the number of items you will take out.

There are 10 digits, and you are choosing sets of 6 digits for your code. Plug the values into the equation:

\text{n = 10, r = 6}

\frac{10!}{(10-6)!} = \frac{10!}{4!} = \boxed{151,200}

There are 151,200 different 6-digit codes possible.
Vlada [557]3 years ago
5 0
Take me the brayniest.i hope it was helpful

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M is the midpoint of CD. Given the coordinates of one endpoint C(10,-5) and the midpoint M(4, 0), find the coordinates of the ot
mash [69]

Answer:

D(-2, 5).

Step-by-step explanation:

We are given that M is the midpoint of CD and that C = (10, -5) and M = (4, 0).

And we want to determine the coordinates of D.

Recall that the midpoint is given by:

\displaystyle M = \left(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2}\right)

Let C(10, -5) be (<em>x</em>₁<em>, y</em>₁) and Point D be (<em>x</em>₂<em>, y</em>₂). The midpoint M is (4, 0). Hence:

\displaystyle (4, 0) = \left(\frac{10+x_2}{2} , \frac{-5+y_2}{2}\right)

This yields two equations:

\displaystyle \frac{x_2 + 10}{2} = 4\text{ and } \frac{y_2 - 5}{2} = 0

Solve for each:

\displaystyle \begin{aligned}x_2 + 10 &= 8 \\ x_2 &= -2 \end{aligned}

And:

\displaystyle \begin{aligned} y_2 -5 &= 0 \\ y_2 &= 5\end{aligned}

In conclusion, Point<em> </em>D = (-2, 5).

5 0
3 years ago
Please solve the problem with steps
Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}

The first one: (\frac{1}{2})^{n-1} is a geometric sequence of first term (a_1) "1" and common ratio (r) " \frac{1}{2} ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2

The second one: \frac{1}{6^{n-1}} is a geometric sequence of first term "1", and common ratio (r) " \frac{1}{6} ". Again, since the common ratio is smaller than one, we can find its infinite sum:

Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}

now we simply combine the results making sure we do the indicated difference: Infinite total sum= 2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}

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3 years ago
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A state park is designed in a circular pattern as shown. A park ranger drives along the circular path from the ranger station to
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HOPE THIS HELPS U
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5 packages of patties = 40

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