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3 years ago

It takes one student 8 hours to wash all of the cars at a school car wash. If 6

Mathematics

1 answer:

hoa [83]3 years ago

4 0

Answer:

C. 1.3 hours

Step-by-step explanation:

It takes one student 8 hours to wash all the cars.

It will take each of 6 students 6 times less to perform same task:

8/6 = 4/3 = 1.3 hours

So the correct choice is C. 1.3 hours

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The soccer team scored three goals in their first game seven goals in their next game in two goals

Butoxors [25]

Answer:

Step-by-step explanation:

3+7+2

12/3

7 0

3 years ago

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 m

Grace [21]

Answer:

0.35

Step-by-step explanation:

we know that

The formula to use for uniform distribution is equal to

$P( X < x) = \frac{x-a}{b-a}$

where,

b is the upper limit of the distribution which is 5 minutes in this case.

a is the lower limit of the distribution which is 0 minutes in this case.

x is the concerned value which is 1.75 minutes in this case.

substitute the given values

$P( X< 1.75) = \frac{1.75-0}{5-0}=0.35$

That means

The probability that a randomly selected passenger has a waiting time less than 1.75 minutes is 0.35

5 0

3 years ago

What is the measurement of the longest line segment in a right rectangular prism that is 26 inches long, 2 inches wide, and 2 in

EastWind [94]

Answer:

$6\sqrt{19} \approx 26.153$ inches.

Step-by-step explanation:

The longest line segment in a right rectangular prism is the diagonal that connects two opposite vertices. On the first diagram attached, the green line segment connecting A and G is one such diagonals. The goal is to find the length of segment $\mathsf{AG}$ .

In this diagram (not to scale,) $\mathsf{AB} = 26$ (length of prism,) $\mathsf{AC} = 2$ (width of prism,) $\mathsf{AE} = 2$ (height of prism.)

Pythagorean Theorem can help find the length of $\mathsf{AG}$ , one of the longest line segments in this prism. However, note that this theorem is intended for right triangles in 2D, not the diagonal in a 3D prism. The workaround is to simply apply this theorem on two different right triangles.

Start by finding the length of line segment $\mathsf{AD}$ . That's the black dotted line in the diagram. In right triangle $\triangle\mathsf{ABD}$ (second diagram,)

Segment $\mathsf{AD}$ is the hypotenuse.
One of the legs of $\triangle\mathsf{ABD}$ is $\mathsf{AB}$ . The length of $\mathsf{AB}$ is $26$ , same as the length of this prism.
Segment $\mathsf{BD}$ is the other leg of this triangle. The length of $\mathsf{BD}$ is $2$ , same as the width of this prism.

Apply the Pythagorean Theorem to right triangle $\triangle\mathsf{ABD}$ to find the length of $\mathsf{AB}$ , the hypotenuse of this triangle:

$\mathsf{AD} = \sqrt{\mathsf{AB}^2 + \mathsf{BD}^2} = \sqrt{26^2 + 2^2}$ .

Consider right triangle $\triangle \mathsf{ADG}$ (third diagram.) In this triangle,

Segment $\mathsf{AG}$ is the hypotenuse, while
$\mathsf{AD}$ and $\mathsf{DG}$ are the two legs.

$\mathsf{AD} = \sqrt{26^2 + 2^2}$ . The length of segment $\mathsf{DG}$ is the same as the height of the rectangular prism, $2$ (inches.) Apply the Pythagorean Theorem to right triangle $\triangle \mathsf{ADG}$ to find the length of the hypotenuse $\mathsf{AG}$ :

$\begin{aligned}\mathsf{AG} &= \sqrt{\mathsf{AD}^2 + \mathsf{GD}^2} \\ &= \sqrt{\left(\sqrt{26^2 + 2^2}\right)^2 + 2^2}\\ &= \sqrt{\left(26^2 + 2^2\right) + 2^2} \\&= 6\sqrt{19} \\&\approx 26.153\end{aligned}$ .