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Tema [17]
3 years ago
8

Two cyclists leave town at the same time on the same road going in  the same direction. Cyclist a is going 6 miles per hour fast

er than cyclist b. After 8 hours, cyclist a has traveled three times the distance as cyclist b. Use the equation 24x= 8(x+6) to find how fast cyclist b was traveling.
Mathematics
2 answers:
tia_tia [17]3 years ago
7 0
24x=8x+48
16x=48
/16  /16

x=3
fgiga [73]3 years ago
6 0

Answer:

3 mph

Step-by-step explanation:

Two cyclists leave town at the same time on the same road going in  the same direction.

Cyclist A is going 6 mph faster than cyclist B

Let speed of cyclist B be x mph

Speed of cyclist A be (x+6) mph

Both cyclist leave town at the same time and traveled 8 hours.  

  • Distance covered by cyclist A in 8 hours= 8(x+6)
  • Distance covered by cyclist B in 8 hours= 8x

After 8 hours cyclist A has traveled 3 times the distance as cyclist B

Therefore, 8(x+6) = 3(8x)

8x + 48 = 24x

24x - 8x = 48

       16x = 48

          x = 3 mph

Hence, The speed of cyclist B was 3 mph

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Answer:

1. The shape of the histogram created with MS Excel is approximately bell shaped and approximately evenly spread about the central (largest counts) values

The shape of the box plot created with MS Excel is approximately evenly distributed

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3. Part A

The 15 numbers between 1  and 20 generated by a random generator are;

1, 8, 8, 15, 4, 18, 11, 6, 17, 1, 18, 15, 10, 12, 11

The measure of center is the mean =  (1+8+8+15+4+18+11+6+17+1+18+15+10+12+11)/15 = 155/15 = 31/3

The measure of spread used is the variance, s² = 32.38

Where, \,s^2 =\dfrac{\sum \left (x_i-\bar x  \right )^{2} }{n - 1}

{\sum \left (x_i-\bar x  \right )^{2} } = 453.333

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s² = 453.\overline 3/(15 - 1) = 32.38

Part B

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Step-by-step explanation:

4 0
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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

5t^{2} =5

t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

7 0
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What is 0.75% of 387
luda_lava [24]

Answer: 2.9025


Step-by-step explanation:

0.75 x 387/100  

= 2.9025

3 0
4 years ago
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