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The quantity demanded x each month of Russo Espresso Makers is 250 when the unit price p is $138. The quantity demanded each mon

Stells [14]

Answer:

(a) $D(q)=\frac{-1}{25} q+148$

(b) $S(q)=\frac{1}{50}q+58$

(c) $p_{*} =88\\\\q_{*} =1500$

Step-by-step explanation:

(a) For the demand equation D(q) we have

P1: 138 Q1: 250

P2: 108 Q2: 1000

We can find m, which is the slope of the demand equation,

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{108-38}{1000-250} =\frac{-30}{750}=\frac{-1}{25}$

and then we find b, which is the point where the curve intersects the y axis.

We can do it by plugging one point and the slope into the line equation form:

$y=mx+b\\\\D(q)=mq+b\\\\138=\frac{-1}{25}(250) +b\\\\138=-10+b\\\\138+10=b=148$

With b: 148 and m: -1/25 we can write our demand equation D(q)

$D(q)=\frac{-1}{25} q+148$

(b) to find the supply equation S(q) we have

P1: 102 Q1: 2200

P2: 102 Q2: 700

Similarly we find m, and b

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{72-102}{700-2200} =\frac{-30}{-1500}=\frac{1}{50}$

$y=mx+b\\\\D(q)=mq+b\\\\72=\frac{1}{50}(700) +b\\\\72=14+b\\\\72-14=b=58\\$

And we can write our Supply equation S(q):

$S(q)=\frac{1}{50}q+58$

(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing D(q)=S(q), which means the demand equals the supply in equilibrium:

$D(q)=S(q)\\\\\frac{-1}{25}q+148=\frac{1}{50}q+58\\\\$

$148-58=\frac{1q}{50} +\frac{1q}{25} \\\\90= \frac{1q}{50} +\frac{2q}{50}\\\\90=\frac{3q}{50}\\ \\q=1500\\\\$

We plug 1500 as q into any equation, in this case S(q) and we get:

$S(q)=\frac{1}{50}q+58\\\\S(1500)=\frac{1}{50}(1500)+58\\\\S(1500)=30+58\\\\S(1500)=88$

Which is the equilibrium price p*.

8 0

3 years ago

The local operations manager for the Internal Revenue Service must decide whether to hire one, two, or three temporary tax exami

Helga [31]

Answer:

If she uses the Laplace criterion, the number of new examiners she will decide to hire is:

c. three

Step-by-step explanation:

a) Data:

Number of Compliance

examiners Low Normal High

One 50 50 50

Two 100 60 20

Three 150 70 -10

b) Outcome Calculations:

Number of Compliance

examiners Low Normal High

One 16.65 (50 *.333) 16.65 (50 *.333) 16.65 (50 *.333) = 50

Two 33.3 (100 *.333) 19.98 (60 *.333) 6.66 (20 *.333) = 60

Three 49.95 (150 *.333) 23.31 (70 *.333) -3.33 (-10 *.333) = 70

c) Decision:

Three has the highest payoff condition and is selected.

d) The Laplace criterion assumes that each compliance state is equally likely to happen. Therefore, it assigns the same weight to each state of compliance. Since there are three states of compliance, we shall assign each state a weight of 0.333. The number of examiners that have the highest payoff condition is three, and therefore, the number "three" is selected.

5 0

3 years ago

Basher collected 4 1/3 basket of peaches and Orchard if each basket holds 21

miv72 [106K]

To do this you times 3 by 4 and add it on to the numerator of the fraction, to turn it into a top heavy fraction:
3*4 = 12
1+12/3 = 13/3

Then you multiply the numerator by 21 to work out what 21 times the fraction is:
13*21/3 = 273/3

Then you can divide 273 by 3 to get the final answer:
273/3 = 91

He will have 91 peaches overall.
Hope this helps! :)

6 0

3 years ago

Write an equation in point-slope form of the line that passes through (-4,1) and (4,3).

SVEN [57.7K]

Step-by-step explanation:

y-y₁=m(x-x₁)

1-(-4)=m(4-3)

5=m

5 0

3 years ago

Given f (x ) = x^2 + 3x + 2 and g (x ) = x + 1, perform the indicated operations.

Nana76 [90]

Answer:

(i) (f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2

Step-by-step explanation:

The given functions are;

f(x) = x² + 3·x + 2

g(x) = x + 1

(i) (f - g)(x) = f(x) - g(x)

∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1

(f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3

(f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = f(x) × g(x)

∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2

(f·g)(x) = x³ + 4·x² + 5·x + 2

7 0

3 years ago