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Lubov Fominskaja [6]
2 years ago
5

How to do this goalookr goalookr

Engineering
1 answer:
Zepler [3.9K]2 years ago
7 0

Well, I am not able to understand your question...

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Write a SELECT statement that returns these column names and data from the Products table: product_name The product_name column
I am Lyosha [343]

Answer:

 SELECT  

   product_name, list_price, discount_percent,    

   ROUND(list_price * (discount_percent / 100), 2) AS discount_amount,

   ROUND(list_price - (discount_percent / 100 * list_price), 2)  AS discount_price  

FROM

Products

ORDER BY (list_price - (discount_percent / 100 * list_price)) DESC

LIMIT 5;

Explanation:

In the above SELECT statement is used to select columns from Products table.

The columns are  product_name,list_price and discount_percent

Next a column discount_amount is selected which is the calculated from previous two columns that are list_price and discount_percent. The discount amount is basically calculated by multiplying list_price with discount_percent/100 and the resultant column is named as discount_amount using Alias which is used to give a temporary name to set of columns or a table.

Next another column discount_price is obtained from previous three columns that are  list_price , discount_percent and discount_amount as: list_price - (discount_percent / 100 * list_price) This as a whole is given a column name discount_price.

FROM is used to refer to a table from which these columns are to be selected. The table name is Product.

The result set is sorted in descending order by discount_price so ORDER BY is used to order the resultant records and DESC is used to sort these records according to the discount_price in descending order.

LIMIT statement is used to extract the records from Product and limit the number of rows returned as a result based on a limit value which is 5 here.

5 0
3 years ago
Question 2 (Multiple Choice Worth 3 points)
ololo11 [35]

Answer:

the car to the right

Explanation:

its in the name the RIGHT of way hope it helps good luck

6 0
3 years ago
Where’s the silicone rubber are made ? In a machine, vulcanization… explain please !!!!! I need help
Vaselesa [24]

Answer: This is done by heating a large volume of quartz sand to temperatures as high as 1800˚C. The result is pure, isolated silicon, which is allowed to cool and then ground into a fine powder. To make silicone, this fine silicon powder is combined with methyl chloride and heated once again.

Explanation:

4 0
3 years ago
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna
Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0
3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
2 years ago
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