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nadezda [96]
1 year ago
9

A construction company plans to build a certain number of apartment buildings and stores on a piece of land. This PPC shows the

combination of projects it can build
1) If the amount of land available to the company increases, the PPC will _____.
a) Shift to the left
b) Shift to the right
c) remain unchained

2) The company realizes it cannot construct any buildings on a portion of the land because it is at risk of a cave-in. In this case, the PPC will _____.
a) Shift to the left
b) Shift to the right
c) remain unchained

Business
2 answers:
VARVARA [1.3K]1 year ago
7 0
1. b) Shift to the right

2. c) Remain unchainged
Bond [772]1 year ago
5 0

Answer 1) Option B) Shift to the right.

Explanation : If the amount of land available to the company increases, the PPC will shift to the right. As the graph indicates, the PPC will grow by shifting on right side as the company is acquiring more land for building purpose.

Answer 2) Option C) Remain Unchanged.

Explanation : The company realizes it cannot construct any buildings on a portion of the land because it is at risk of a cave-in.

In this case, the PPC will remain unchanged. When the company realizes that no construction can be done on the portion of land because of its hollowness the PPC will remain to be undisturbed.

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The journal entry every year will be 
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Three years ago, law school admits deciding whether or not to attend the schools they were admitted to typically underestimated
V125BC [204]

Answer:

Limited Supply of lawyers will lead to increase in Lawyer Wages / Salaries

Explanation:

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Less certain a cash flow, the ________ the risk, and ________ the present value of the cash flow. higher; lower lower; lower hig
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Answer:

probability  = 0.3557

Explanation:

given data

young adults ages =  20 to 39

skip eating breakfast p = 0.238

random sample of size n = 500

to find out

we find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122

solution

we use here Normal Approximation to Binomial Distribution

so first consider random variable = x

so

x~ Bin (n,p)   .............1

and here Normal Approximation will be

x~ Normal Approx (np, npq)    .................2

so it will be

x~ (500, 0.238)  

as here we know q will be

q = 1 - p

q = 1 - 0.238

q = 0.762    .............3

so

here x~ Normal Approx (119, 90.678)

and now we get P(X > 122)

so

We will convert it to Z by as that

z = \frac{x-\mu}{\sigma}     ................4

and here

mean  \mu = np

and standard deviation \sigma =  \sqrt{npq}

so here for P(X > 122)

P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})     ............5

and it is  P(Z>0.37)

so

probability  = 1 - P(Z<0.37)

now we use here z table for value

probability  = 1-0.6443

probability  = 0.3557

7 0
1 year ago
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