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Mnenie [13.5K]
3 years ago
11

And

Mathematics
1 answer:
kramer3 years ago
5 0

Answer:

can you send a picture please so i can see it?

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A clock is positioned on an auditorium wall with its center 9 ft above the floor. The second hand on the clock is 10 inches long
oee [108]

Answer:

B : y=5/6cos(pi/30x)+9

Step-by-step explanation:

Edge 2020

7 0
3 years ago
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If the sum of difference of 13 and 4 divided 64 and 8 is subtracted from the result. Find that number. Mathematical equation plz
natka813 [3]

Answer:

13+4=17,17/64=3-8=-5

Step-by-step explanation:

this is the solution

3 0
2 years ago
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Solve the following inequality. |3n-2|-2<1
disa [49]

Answer: \frac{-1}{3} < n < \frac{5}{3}

<u>Step-by-step explanation:</u>

| 3n - 2 | - 2 < 1

<u>               +2 </u> <u>+2 </u>

| 3n - 2 |       < 3

3n - 2 < 3     and     3n - 2 > -3

<u>      +2 </u> <u>+2   </u>              <u>     +2 </u>  <u> +2 </u>

3n       < 5      and     3n      > -1

       n <  \frac{5}{3}       and         n > \frac{-1}{3}

\frac{-1}{3} < n < \frac{5}{3}

Interval Notation:  (\frac{-1}{3},\frac{5}{3})

Graph:  \frac{-1}{3} o--------------------o \frac{5}{3}


7 0
3 years ago
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A bucket contains one green block, one red block, and two yellow blocks. You choose one block from the bucket.
nalin [4]
A. 1/4; 2/4=1/2; 1/4
b. 1
c. 3/4. There are 4 blocks and 3 are not red
d. 1/4+3/4= 4/4=1


8 0
2 years ago
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
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