Find the value of the combination. 7 C 2 a.21 b.42 c.49

Alenkinab [10]

sorry this is late!!

#1 - A circular grid like this one can be helpful for performing dilations

#2 - To perform a dilation, we need a (In order to perform a dilation, students will need to know the center of dilation (which can be communicated using the coordinate grid), the coordinates of the polygon that they are dilating (also communicated using the coordinate grid), and the scale factor.)

5 0

3 years ago

Suppose a new car is purchased for $46,357 and depreciates by 22% over the first year of ownership. If the car is driven 12,170

miss Akunina [59]

Answer:check the pic

Step-by-step explanation:

3 0

3 years ago

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Solve this equation 5x+2=3x-20-1x

Rainbow [258]

$5x+2=3x-20-1x\\5x+2=2x-20\ \ \ \ |subtract\ 2\ from\ both\ sides\\5x=2x-22\ \ \ \ |subtract\ 2x\ from\ both\ sides\\3x=-22\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=-\frac{22}{3}}\Rightarrow\boxed{x=-7\frac{1}{3}}$

6 0

3 years ago

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An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same

elena-s [515]

Answer:

Approximately $0.15$ ( $360 / 2401$ .) (Assume that the choices of the $5$ passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all $7$ floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these $5$ passenger could choose from any one of the $7$ floors. There would be a total of $7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$ unique ways for these $5\!$ passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the $7$ floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only $(7 - 1) = 6$ floors.

Likewise, the third passenger would have to choose from only $(7 - 2) = 5$ floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only $(7 \times 6 \times 5 \times 4 \times 3)$ unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the $7$ floors. Each of these $7^{5}$ combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

$\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}$ .

5 0

2 years ago

Which graphs show continuous data? Select each correct answer.

Nikolay [14]

Answer:

I say the first graph in the first pic

Also,the 2nd graph in the 2nd pic

7 0

3 years ago

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