Answer:
The correct answer is option C.
Explanation:
Market failure refers to the situation when the market is not able to efficiently allocate resources and the government has to intervene. Market failure generally happens because of the presence of externalities.
When the marginal social cost is greater than the ability and willingness to pay, the market will fail to optimally allocate resources. The government, as a result, will intervene.
The government will use vouchers which will cause the marginal private benefit curve to shift upwards by the size of the per-unit voucher.
The journal entry for the issuance of the stock for issue of 40 shares at a par value of $20, will affect a credit to Common Stock for $800.
<h3>What is a journal entry?</h3>
The process of maintenance of systematic and chronological records of financial transactions during a given financial period is known as a journal entry.
Hence, option C holds true regarding the journal entry.
Learn more about journal entry here:
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Answer:
a. Sale of Common Stock.
Classification: Financing activity
b. Sale of Land
Classification: Investing activity
c. Purchase of Treasury Stock
Classification: Financing activity
d. Merchandise Sales
Classification: Operating activity
e. Issuance of a long-term note payable
Classification: Financing activity
f. Purchase of merchandise
Classification: Operating activity
g. Repayment of note payable
Classification:
Financing activity
h. Employee salaries
Classification: Operating activity
i. Sale of equipment at a gain.
Classification: Investing activity
j. Issuance of bonds
Classification: Operating activity
Answer:
True
Explanation:
Jeffrey Pfeffer a Professor of Organizational Behavior and Robert I. Sutton a Professor of Management Science and Engineering both Stanford’s Graduate School of Business gave an approach to decision making in an organization as an action that is driven by hard facts rather than half truths.
Answer:
(a) <em>z</em> = 0.53
(b) <em>z</em> -0.67
(c) <em>z</em> = -0.84
(d) <em>z</em> = 0.26
Explanation:
A standard normal distribution has mean 0 and standard deviation 1.
(a)
Compute the value of <em>z</em> for P (<em>Z</em> < <em>z</em>) = 0.70 as follows:
Consider the<em> </em>standard normal table for the value of <em>z.</em>
The value of <em>z</em> is 0.53.
(b)
Compute the value of <em>z</em> for P (<em>Z</em> < <em>z</em>) = 0.25 as follows:
Consider the<em> </em>standard normal table for the value of <em>z.</em>
The value of <em>z</em> is -0.67.
(c)
Compute the value of <em>z</em> for P (<em>Z</em> > <em>z</em>) = 0.20 as follows:
P (Z > z) = 0.20
1 - P (Z < z) = 0.20
P (Z < z) = 0.80
Consider the<em> </em>standard normal table for the value of <em>z.</em>
The value of <em>z</em> is -0.84.
(d)
Compute the value of <em>z</em> for P (<em>Z</em> > <em>z</em>) = 0.60 as follows:
P (Z > z) = 0.60
1 - P (Z < z) = 0.60
P (Z < z) = 0.40
Consider the<em> </em>standard normal table for the value of <em>z.</em>
The value of <em>z</em> is 0.26.
