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3 years ago

HELLLLPPPPPPPPPPPPPPPPPPPPPPP PLS

Mathematics

2 answers:

Sladkaya [172]3 years ago

8 0

I think C. and if you need to solve it x=2

Tju [1.3M]3 years ago

7 0

Answer:

Step-by-step explanation:

x is the hours that the job is being done, so we can use 8x and 11x. The plus 6 comes from the +6 he earns from the edging done.

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-2 3/4 ÷ -1/7= <br>help

Hitman42 [59]

The answer is 19 $\frac{1}{4}$

First you must set up the equation

$-2\frac{3}{4}$ ÷ $-\frac{1}{7}$

First, you must make the mixed number a improper fraction

$-\frac{11}{4}$ ÷ $-\frac{1}{7}$

Then you must divide

$\frac{77}{4}$

Finally, simplify

19 $\frac{1}{4}$

3 0

3 years ago

Q2) The following travel times were measured for vehicles traversing a 2,000 ft. segment of an arterial: Vehicle Travel Time (s)

adelina 88 [10]

Answer:

TMS = 45.78 m/s

SMS = 45.66 m/s

Step-by-step explanation:

Vehicle Travel Time (s)

1 40.5

2 44.2

3 41.7

4 47.3

5 46.5

6 41.9

7 43.0

8 47.0

9 42.6

10 43.3

Time mean speed and space mean speed are parameters used to express the speed of a group of vehicles in traffic flow.

Time mean speed is given as the average speed of all the cars

The speed for each vehicle is calculated by dividing the (2000 ft) distance by the time taken by each vehicle to complete that segment of the journey.

Let vehicle = V

Travel time = time

Speed = s

V | Time | speed (ft/s)

1 | 40.5 | 49.383

2 | 44.2 | 45.249

3 | 41.7 | 47.962

4 | 47.3 | 42.283

5 | 46.5 | 43.011

6 | 41.9 | 47.733

7 | 43.0 | 46.512

8 | 47.0 | 42.553

9 | 42.6 | 46.948

10| 43.3 | 46.189

TMS = (Σsᵢ)/N

Σsᵢ = (49.383+45.249+47.962+42.283+43.011+47.733+46.512+42.553+46.948+46.189)

Σsᵢ = 457.823

N = number of vehicles = 10

TMS = 457.823 ÷ 10 = 45.7823 = 45.782 m/s

b) Space mean speed is given as

SMS = N ÷ [Σ (1/sᵢ)]

V | Time | speed | (1/sᵢ)

1 | 40.5 | 49.383 | 0.02025

2 | 44.2 | 45.249 | 0.02210

3 | 41.7 | 47.962 | 0.02085

4 | 47.3 | 42.283 | 0.02365

5 | 46.5 | 43.011 | 0.02325

6 | 41.9 | 47.733 | 0.02095

7 | 43.0 | 46.512 | 0.02150

8 | 47.0 | 42.553 | 0.02350

9 | 42.6 | 46.948 | 0.02130

10| 43.3 | 46.189 | 0.02165

Σ (1/sᵢ) = (0.02025+0.02210+0.02085+0.02365+0.02325+0.02095+0.02150+0.02350+0.02130+0.02165)

Σ (1/sᵢ) = 0.219

SMS = 10 ÷ [Σ (1/sᵢ)] = 10 ÷ 0.219

SMS = 45.662 m/s

Hope this Helps!!!

4 0

3 years ago

What is the area of the triangle below (in square units)?<br> 24in<br> 2 in

UNO [17]

Answer:

24 in

Step-by-step explanation:

Area of triangle = bh/2

= (2 * 24)/2

= 24 in

8 0

3 years ago

Read 2 more answers

4,284 passengers traveling by the intercity bus every day, the intercity bus operates 6 hours every day, find the number of pass

Lemur [1.5K]

Answer:

714

Step-by-step explanation:

So lets go over what we know.

The bus only goes for 6 hours each day.

So lets just think of a full day as 6 hours.

Each day(6 hours), 4284 passangers ride the bus.

To find passangers per hour, we must divide the total passangers by the total hours.

This is 4284(total passangers) divided by 6(total hours):

4284/6

714

So each hour, there are 714 passangers.

Hope this helps!

7 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago