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Ivanshal [37]
3 years ago
15

What was the opening price of Dow Jones Industrial Average on July 06,2017 in the format of XXXXX.XX?

Engineering
1 answer:
Rashid [163]3 years ago
3 0

Answer:

hello your question is poorly written hence I will give you a general answer

answer :

Hence the opening price =  140 - ( 14 * 3 )

                                           =  140 - 42 = 98.00

Explanation:

To determine the opening price we will subtract or add the percentage Increase or decrease of the Dow Jones from the Final price of the Dow Jones

Lets assume :

Dow Jones Increased by 30 %

Dow Jones final price = 140

Hence the opening price =  140 - ( 14 * 3 )

                                           =  140 - 42 = 98.00

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How to plot 0.45 gradation chart for sieve analysis ?
LekaFEV [45]

Answer:

532235w3r35w3r

Explanation:

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5 0
2 years ago
Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum
ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

<em>Make the assumption Base continuous flow velocity (BFFS)= 65 mph. </em>

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

6 0
3 years ago
Read 2 more answers
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
If superheated water vapor at 30 MPa iscooled at ​constant pressure​, it will eventually become saturated vapor, and with suffic
nirvana33 [79]

Answer:

False.

Explanation:

False. The pressure is above pressure at critical point (22.064 MPa.), the limit where pressure can prevent boiling.

3 0
3 years ago
What else will change, if you change the point of view
JulijaS [17]

Answer:

We would need background context,

Explanation:

Then I would be happy to help!

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