Choose the angle whose measure is greater than 90 degrees but less than 180 degrees.

1 answer:

4 0

B) An obtuse angle would have a measure between 90 and 180.

Acute triangles would have a measure less than 90, right angles would have a measure of exactly 90, and straight angles would have a measure of exactly 180.
Hope this helps!

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Jonah's and his aunt's houses are 8,750 meters apart. Jonah's walking speed is 5 kilometers per hour. How long will it take him

umka2103 [35]

8750 Meters = 8.75 Kilometers
8.75 Kilometers / 5 Kilometers Per Hour = 1.75 Hours = 1 Hour and 45 Minutes

Hope this helps.

5 0

3 years ago

I need help with problem

pychu [463]

Answer:

f(x)=4x(x-3)

Vertex=3/2, -9

Step-by-step explanation:

First, factorize it.

That is equal to 4x(x-3)

Therefore, x intercepts are 0 and 3, as at least one of them need to be 0.

To find the vertex, first find the x of the vertex, which you can do by finding the average of the x intercepts, so it will be 3/2

Then, find the y, by making x in the original equation equal 3/2. This will give us -9 as the y.

The vertex will by 3/2, -9

4 0

2 years ago

wheelchair ramps for access to public buildings are allowed a maximum of one inch of vertical increase for every one foot of hor

AveGali [126]

Length x width = area
but
1feet=12inches
8 x 12=96
10x96=960

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.