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postnew [5]
3 years ago
8

Subtract. 4 1/5 -2 9/10

Mathematics
2 answers:
Vikki [24]3 years ago
6 0
Multiply by 2 on the numerator and the denominator of 4 1/5, we get 4 2/10, which is 4.2. 2 9/10 is 2.9. We subtract 4.2 - 2.9 to get 1.3, which, as a fraction, is 1 3/10.
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Hope this helps!
==jding713==
Lerok [7]3 years ago
3 0
4 1/5 - 2 9/10 = 4 2/10 - 2 9/10 = 3 12/12 - 2 9/10 = 1 3/10. Hope this helped
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4 years ago
Is (2, 2) a solution of y < 4x – 6?
Roman55 [17]

Answer:

No

Step-by-step explanation:

To see if (2,2) is a solution, substitute them into the inequality and see if it is true.

2 is our x value, and 2 is also our y value

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2< 4(2)– 6

2 < 8 – 6

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This statement is false. 2 is not less than 2, so (2,2) is not a solution to this equation.

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3 years ago
Suppose account numbers for an internet service provider consist of alpha-numeric characters. The first character is a digit (0
Amiraneli [1.4K]

Answer:

20,442,240 account numbers are possible.

Step-by-step explanation:

The first character has 10 possible values.

The next 2 has 26 * 26 = 676 possible combinations.

The last 4 has 9*8*7*6 = 3024 possible combinations.

So the answer is 10 * 676 * 3024

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6 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
I need help. Please ​
GREYUIT [131]

Answer:

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Step-by-step explanation:

A=1/2(l*w)

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A=1/2(10)

A=5

3 0
3 years ago
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