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olga_2 [115]
3 years ago
11

Using the midpoint method, calculate the price elasticity of demand of Good Z using the following information: When the price of

good Z is $10 (P1), the quantity demanded of good Z is 85 units (Q1). When the price of good Z rises to $15 (P2), the quantity demanded of good Z falls to 60 units (Q2).
(A) The price elasticity of demand for good Z = 1.90.
(B) The price elasticity of demand for good Z = 0.86.
(C) The price elasticity of demand for good Z =0.76.
(D) The price elasticity of demand for good Z = 0.52.
Mathematics
1 answer:
exis [7]3 years ago
6 0

Answer: (B) The price elasticity of demand for good Z = 0.86

Step-by-step explanation:

The formula for determining elasticity of demand by using the midpoint method is

(Q2 - Q1)/[(Q2 + Q1)/2] / (P2 - P1)/[(P2 + P1)/2]

Where

P1 is the initial price of the item.

P2 is the final price of the item.

Q1 is the initial quantity demanded for the item.

Q2 is the final quantity demanded for the item.

From the information given,

P1 = 10

P2 = 15

Q1 = 85

Q2 = 60

The price elasticity of demand for good Z = (60 - 85)/[(60 + 85)/2] / (15 - 10)/[(15 + 10)/2]

= (-25/72.5) / (5/12.5) = -25/72.5 × 12.5/5

= - 312.5/362.5 = - 0.86

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1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

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k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

when,

k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

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k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

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