The following graph shows the market for sprockets in a local hardware store.
a) identify whether the price of sprockets from $4 - $3 is elastic, unit elastic, or inelastic. Explain.
b) what is the price of sprockets that maximize the hardware store’s revenue? Explain.
c) The price of sprockets is $1. if the hardware store wants to increase its revenue, how should it change the price? Explain.
d) using the midpoint formula, calculate the price elasticity of demand for sprockets when price increases from $2 to $3. Identify the coefficient as elastic, unit elastic, or inelastic.
e) using the midpoint formula, calculate the price elasticity of demand for sprockets when price decreases from $5 to $4. Identify the coefficient as elastic, unit elastic, or inelastic.
f) based on the scenario in part E, how does the decrease in price affect the hardware store’s revenue?
Answers
It is correct to state that the price for sprockets from $4 to $3 is inelastic. See the explanation below.
What is the calculation supporting the answer above?The formula for price elasticity is:
Price Elasticity of Demand = Percentage change in quantity / Percentage change in price.
Percentage change in quantity (for price drop of 4$ to $3) = increase ÷ original number × 100.
= (2/18) * 100
= 11.11%
Percentage change in price drop from 4$ to $3) = decrease ÷ original number × 100.
= (1/4) * 100
= 25%
Hence
Price Elasticity of Demand = 11.11%/25%
PED = 0.4444
Hence the demand is inelastic for the change in price from $4 to $3. This is because the price Elasticity is less than 1.
Because we don't have the function of the graph, we must find the revenue for all the given points.
Recall that Revenue is given as:
Price x Quantity; Hence,
From the above, it is clear that the price that maximizes the revenue the store is $4. This is because, that yields the highest revenue.
Using the midpoint formula the price elasticity of demand for sprockets when price increases for $2 to $3 is?The mid-point formula is given as:
Price Elasticity = A/B where:
A = [(Q2-Q1)/((Q2+Q1)/2)]
B = [(P2-P1)/((P2+P1)/2)]
Q2 = Final Quantity = 18
Q1 = Initial Quantity = 20
P2 = Final Price = $3
P1 = Initial Price = $2
A = [(18-20)/(18+20)/2]]
= -0.02631578947
B = [(3-2)/((2+3)/2)]
= 0.4
PED = -0.026/0.4
PED = -0.065
From the foregoing, the Coefficient is Negative and it is also less than Zero. It means that the good is inferior and the demand for it is inelastic.
Using the midpoint formula the price elasticity of demand for sprockets when price decreases for $5 to $4 is?Using the same formula above we have:
A = [(Q2-Q1)/((Q2+Q1)/2)]
B = [(P2-P1)/((P2+P1)/2)]
Q2 = Final Quantity = 16
Q1 = Initial Quantity = 10
P2 = Final Price = $4
P1 = Initial Price = $5
A = [(16-10)/(16+10)/2]]
= 0.11538461538
B = [(4-5)/(4+5)/2]]
= -0.05555555555
PED = 0.11538461538/-0.05555555555
PED = -2.08
The coefficient here is also negative but above 1. It means that the good is inferior and the price and quantity demanded are inversely related.
However, the value (2) is greater than 1, thus, the demand is elastic. This means there is a strong reaction to quantity demanded when price increases or decreases.
Based on the scenario in part E, how does the decrease in price affect the hardware store’s revenue?
It is to be noted that decreasing the price as given in part E will trigger a drop in sales and a directly proportional reduction in profit. The reason for this is because, as seen in the analysis for E above the product is inferior.
When people perceive a product as inferior, a drop in price will create a substitutionary effect where they opt for a substitute good that is slightly higher.
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Answer:
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These include, land, labor, people, minerals
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N2 + 3H2 2NH3
If N2 is reacting at a rate of –0.40 M/s, what is the rate by which NH3 is being produced? (Also, show calculations.)
-0.80 M/s
+0.40 M/s
+0.80 M/s
+1.20 M/s
Answers
Answer:
+0.8M/s
Explanation:
We are given that
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
Rate of reacting of N2, d/dt([N2])=-0.4M/s
We have to find the rate by which NH3 is being produced.
We know that
[tex]-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}[/tex]
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[tex]\frac{d[NH_3]}{dt}=0.4\times 2[/tex]
[tex]\frac{d[NH_3]}{dt}=0.8M/s[/tex]
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Answer:
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Explanation:
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Time taken t
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Answer:
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Answer:
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I just did this on edge. Brainliest?
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Answers
Answer:
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Answer:
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This is the correct answer of edge as I got a 2/2 (100%) on this section of the test. There is further proof that I am correct in the file attached.
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ty
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