HELP ME PLEASE RN

Write a SELECT statement that returns one row for each customer that has orders with these columns: The email_address column fro

Dafna11 [192]

Answer:

The statement is as follows:

Explanation:

We had better avoid such correlated subqueries by instead using aggregation with GROUP BY:

SELECT

c.email_address,

COUNT(DISTINCT o.order_id) AS num_orders,

COALESCE(SUM(oi.quantity * (oi.item_price - oi.discount_amount)), 0) AS total_amount

FROM customers c

LEFT JOIN orders o

ON c.customer_id = o.customer_id

INNER JOIN order_items oi

ON o.order_id = oi.order_id

GROUP BY

c.customer_id,

c.email_address;

8 0

3 years ago

Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M

weqwewe [10]

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

$500 = \frac{P}{\frac{\pi}{4} 5^2}$

solving for P

P = 9817.47 N

for Aluminium bar

$150 = \frac{P}{\frac{\pi}{4} 10^2}$

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

6 0

3 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

3 0

3 years ago

Use differentials to estimate the amount of metal in a closed cylindrical can that is 30 cm high and 6 cm in diameter if the met

PilotLPTM [1.2K]

Answer:

$dV=113.55 \ cm^3$

Explanation:

<u>The Differential of Multivariable Functions</u>

Given a multivariable function V(r,h), the total differential of V is computed by

$dV=\displaystyle \frac{\partial V}{\partial r} dr+\frac{\partial V}{\partial h} dh$

The volume of a cylinder of radius r and height h is

$V=\pi\cdot r^2\cdot h$

Let's compute the partial derivatives

$\frac{\partial V}{\partial r}=2\pi rh$

$\displaystyle \frac{\partial V}{\partial h}=\pi r^2$

The total differential is

$dV=(2\pi rh)dr+(\pi r^2)dh$

The differential dr is approximated to $\Delta r$ and the differential dh is approximated to $\Delta h$ . We can see the increment of radius is the thick of the metal in the sides, and the increment of the height is the thick of the metal in the top and bottom. Thus dr=0.05 cm, dr=0.2 cm