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goldfiish [28.3K]
3 years ago
11

The price of cookies was increased from $4 to $5, the quantity demanded by stores went down 50%, what is price elasticity of dem

and?
Mathematics
1 answer:
Ede4ka [16]3 years ago
7 0
Assume the normal quantity of cookies is x.

The demanded quantity by the stores now:
0.5x
The previous cost is: $4 * 0.5x = $2x

The cost of demanded quantity by the stores now: $5 * 0.5x = $2.5x

The price elasticity of demand: $2.5x - $2x = $0.5x

Hope that helps you
You might be interested in
FGH~CDE, solve for x <br><br> 13<br><br> 11<br><br> 4<br><br> 14
babunello [35]

Answer:

Choice 2) x = 11

Step-by-step explanation:

7/14 = 8/(x + 5)

8(14) = 7(x + 5)

112 = 7x + 35

7x = 77

X = 11

5 0
3 years ago
You increase the size of a computer screen display by 20%. Then you decrease it by 20%. What is the size of the computer screen
lutik1710 [3]

Answer: The size of computer display after 20 % increase and then 20 % decrease= 96 % of the original size.


Step-by-step explanation:

Let the original size of computer be 100x , then if it increases by 20 %

then the new size of the computer=100x +20% of 100x

=100x+\frac{20}{100}\times100x\\=100x+20x\\=120x

The size of computer after 20% increase= 120x

If size of computer is decreased by 20 % then

The new size of computer= 120x - 20% of 120 x

=120x-\frac{20}{100}\times120x\\=120x-24x\\=96x

On comparing it to the original size , we get

\frac{96x}{100x}\times100=96\%

Thus, the size of computer display after 20 % increase and then 20 % decrease= 96 % of the original size.

5 0
3 years ago
Read 2 more answers
Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be
Vedmedyk [2.9K]

Answer: 63 defective widgets

Step-by-step explanation:

Given that the proportion should not exceed 5%, that is:

p< or = 5%.

So we take p = 5% = 0.05

q = 1 - 0.05 = 0.095

Where q is the proportion of non-defective

We need to calculate the standard error (standard deviation)

S = √pq/n

Where n = 1024

S = √(0.05 × 0.095)/1024

S = 0.00681

Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.

UCL = p + Za/2(n-1) × S

Where a is alpha of confidence interval = 100 -90 = 10%

a/2 = 5% = 0.05

UCL = p + Z (0.05) × 0.00681

Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.

Z a/2 = 1.64

UCL = 0.05 + 1.64 × 0.00681

UCL = 0.0612

Since the UCL in this case is a measure of proportion of defective widgets

Maximum defective widgets = 0.0612 ×1024 = 63

Alternatively

UCL = p + 3√pq/n

= 0.05 + 3(0.00681)

= 0.05 + 0.02043 = 0.07043

UCL =0.07043

Max. Number of widgets = 0.07043 × 1024

= 72

7 0
3 years ago
Can someone PLEASE show me how to do this! If f (x) = x^2 + 3x +2/ x +3, then f’ (x) =
coldgirl [10]

Answer:

c

Step-by-step explanation:

The equation =( denominater * derivative of numerator - numerator * derivative of denominator) / denominator ^2

so the qstn is (x^2 + 3x +2) / (x+3)

apply the values as the above eqtn states

ie,[ (x+3) * derivative of (x^2 +3x + 2)] - [( x^2 +3x + 2) *derivative of (x+3)] /

(x+3)^2

derivative of numerator, (x^2 +3x + 2) is 2x+3

" of denominator, (x+3) is 1

so we get

[(x+3)* (2x + 3 ) - (x^2 +3x + 2) *1 ] / (x+3)^2

open the brackets

[ 2x^2 + 3x + 6x + 9 - x^2 +3x + 2 ] / (x+3)^2

subtract similar terms and we get the final answer in option c

7 0
3 years ago
Divide 15 into thirds and subtract from 14 divided in half
Maurinko [17]
-2, because 15 divided by 3 is 5 and 14 divided by 2 is 7 so 5 minus 7 is -2
8 0
3 years ago
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