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A farmer wants to buy a 10 kg bag of fertilizer (organic soil). They have the choice between two merchants. Merchant A sells the

sergejj [24]

Answer:

Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.

Explanation:

Which merchant has the better deal means which merchant offers the farmer a better deal.

For Merchant A, 10 kg bag = $10

meaning it contains a real 10 kg bag of dry fertilizer which the farmer can use without losing any Kg to drying.

While for Merchant B, 10 kg bag = $8

where the 10kg = 80% dry fertilizer + 20% water content

But the farmer can only use the solid constituents of the bag which means,

Merchant B is giving 80/100 x 10Kg of dry fertilizer for $8

That is, 8kg for $8

Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.

5 0

2 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

Anettt [7]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

Pressure difference given as

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$

7 0

3 years ago

Thank you very much this helped a lot

Len [333]

Answer:

Your welcome i guess

Explanation:

7 0

2 years ago

Read 2 more answers

one airline limits the size of carry-on luggage to a volume of 40,000 cm. A passenger has a carry-on that has an area of 1,960 c

PolarNik [594]

<u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u>

Explanation:

<h2>Given:</h2>

Acceptable volume = 40,000 cm³

Area of luggage = 1,960 cm²

Height of luggage = 23 cm

<h2>Question:</h2>

Is the passenger's luggeage ok to carry onto the airplane

<h2>Equation:</h2>

V = l x w x h

or we can use

V = A x h

Since A = l x w

where: V - volume

A - area

l - length

w - width

h - height

<h2>Solution:</h2>

V = A x h

V = ( 1,960 cm²)(23 cm)

V = 45,080 cm³

45,080 cm³ is greater than the acceptable volume 40,000 cm³

<h2>Final Answer:</h2><h3><u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u></h3><h3 />

7 0

2 years ago

Past evidence shows that when a customer complains of an out-of-orderphone there is an 8% chance that the problem is with the in

Alexxandr [17]

Answer:

a. The expected number of failures due to a problem with inside wiring is 8 failures

b. The probability that at least 10 failures are due to inside wiring is approximately 0.176

c. It will be not unusual

Explanation:

The probability that the problem of an out of order is the inside wiring, P(x) = 8%

The number of complaints in a month period, x = 100

a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)

∴ E(x) = 100 × 8% = 8

The expected number of failures due to a problem with inside wiring, E(x) = 8 failures

b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;

The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92

P = $_nC_r$ · $P^r$ · $q^{n-r}$

P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657

P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325

P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793

P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616

P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593

P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653

P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561

P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735

P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516

P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125

P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362

∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464

The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536

The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176

c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908

Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring

6 0

3 years ago

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