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2 years ago

14

Select the correct answer from the drop-down menu. A box contains shirts in two different colors and two different sizes. The nu

mbers of shirts of each color and size are given in the table. Shirt Color Size Large Medium Total Red 42 48 90 Blue 35 40 75 Total 77 88 165 From the data given in the table, we can infer that .

2 answers:

kogti [31]2 years ago

8 0

Answer:

Plato Users: its P(blue shirt l large shirt) = P(blue shirt)

Explanation:

yKpoI14uk [10]2 years ago

7 0

Answer:

P( red | large shirt ) ≠ P( large shirt)

P( Blue | large shirt ) = P( blue shirt)

P( shirt is medium and blue ) ≠ P( medium shirt)

P( large shirt | red shirt ) ≠ P( red shirt)

Explanation:

P( red | large shirt ) = 42/77 = 0.5454 : P( large shirt)= 77/165 = 0.467

P( red | large shirt ) ≠ P( large shirt)

P( Blue | large shirt ) = 35/77 = 0.4545 : P( blue shirt)= 75/165 = 0.4545

P( Blue | large shirt ) = P( blue shirt)

P( Shirt is medium and blue ) = 40/77 = 0.2424 : P( medium shirt)= 88/165 = 0.5333

P( shirt is medium and blue ) ≠ P( medium shirt)

P( large shirt | red shirt) = 42/90 = 0.4667 : P( red shirt)= 90/165 = 0.5454

P( large shirt | red shirt ) ≠ P( red shirt)

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

payback period = $\frac{32 + 66}{520.34}$

payback period = 0.188 year

8 0

2 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

2 years ago

A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.

Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0

1 year ago

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0

2 years ago

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°

Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

Pressure $P_{2}= 70.0\ kPa$

Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

$V_{2}=\dfrac{101.325\times2\times280}{293\times70}$

$V_{2}=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

5 0

2 years ago