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Kay [80]
3 years ago
8

14tx=46;x=32 what's the answer?​

Mathematics
2 answers:
shepuryov [24]3 years ago
6 0

Answer:

t=23/224   x=32

Step-by-step explanation:

[14tx=46]

x=32

14t  times 32 =46

isolate t for 14 t 32=46

t=23/224

pickupchik [31]3 years ago
3 0

First plug 32 in for x 14(32)t=46

Now solve for t

448t=46

T=46/448 which can be simplified to 23/224

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30 POINTS FOR THE GENUIS WHO CAN ANSWER THIS!!
bearhunter [10]

Answer:

Answers are below in bold

Step-by-step explanation:

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A = 1/2(8)(6)         Multiply

A = 1/2(48)           Multiply

A = 12cm²           Area of each triangular base

2) A = L x W        Use this equation to find the area of the bottom rectangular face

A = 20 x 8          Multiply

A = 160 cm²       Area of the bottom rectangular face

3) A = L x W        Use this equation to find the area of the back rectangular face

A = 20 x 6          Multiply

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4) A = L x W        Use this equation to find the area of the sloped rectangular face

A = 20 x 10         Multiply

A = 200 cm²      Area of the sloped rectangular face

5) To find the total surface area of the triangular prism, add together all of the numbers.

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6 0
3 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.

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Answer:

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<em>I hope this helps!!</em>

<em>- Kay :)</em>

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