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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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4 0
3 years ago
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The dry unit weight of a soil sample is 14.8 kN/m3.
Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(c) Degree of saturation

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0
4 years ago
Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle
Vinvika [58]

Answer:

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal  

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside  temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal  = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P = \frac{7800008400 \ Watt-Hour}{3650 \ Hours}  = 2136988.6 \ W

Rate of heat transfer, P = \frac{K*A *\delta T}{L}

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

P = \frac{KA(T_2-T_1)}{L} \\\\L =  \frac{KA(T_2-T_1)}{P}\\\\L =  \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m

L = 6.23 cm

Therefore, the thickness of the material is 6.23 cm

8 0
3 years ago
Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w
ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0
3 years ago
To become familiar with the general equations of plane strain used for determining in-plane principal strain, maximum in-plane s
lukranit [14]

Answer:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

Explanation:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

The explanation is shown in the attachment. I hope i have been able to help.

3 0
3 years ago
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