Answer:
$205,000
Explanation:
Total liabilities=current liabilities+long-term liabilities
total liabilities=$150,000+$220,000
total liabilities=$370,000
total owners'equity plus liabilities=$320,000+$370,000=$690,000
long-term assets+current assets=liabilities+owners'equity
long-term assets=$485,000
current assets are unknown
liabilities+owners'equity=$690,000
let CA represent current assets
$485,000+CA=$690,000
CA=$690,000-$485,000
CA=$205,000
Answer:
0.25
Explanation:
Given the following outcomes,
- Outcome 1: probability (P) = 0.25, return (R) = 0.10
- Outcome 2: P = 0.50, R = 0.25
- Outcome 3: P = 0.25, R = 0.40
The expected return on the investment
= 
= (0.25 * 0.10) + (0.50 * 0.25) + (0.25 * 0.40)
= 0.025 +0.125 + 0.100
Expected return = 0.25.
Answer:E. The human resource department did not promote Fatima because they thought that she would not be able to travel as frequently as the job requred because she has two young children.
Explanation: This action constituted a breach of the Civil Rights Act.
Answer:
Full question: <em>On their birthdays, employees at a large company are permitted to take a 60-minute lunch break instead of the usual 30 minutes. Data were obtained from 10 randomly selected company employees on the amount of time that each actually took for lunch on his or her birthday. The company wishes to investigate whether these data provide convincing evidence that the mean time is greater than 60 minutes. Of the following, which information would NOT be expected to be a part of the process of correctly conducting a hypothesis test to investigate the question, at the 0.05 level of significance?</em>
<em>Answe</em><em>r: Since that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes. </em>
Explanation:
<em>From the given question let us recall the following statements:</em>
<em>Employees at a large company are permitted to take a 60-minute Lunch break instead of the 30 minutes.</em>
<em>Data was gotten from = 10 randomly selected company employees on the amount of time that each actually took for lunch on his or her birthday</em>
<em>Given that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes.</em>
<em>The company tries to investigate the data to know that the mean is greater than 60 minutes</em>
<em>the next step is to find the process of correctly conducting a hypothesis test to investigate the question, at the 0.05 level of significance</em>
<em>Therefore,</em>
<em>Since that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes. </em>
<em>Or</em>
<em>The P-value> 0.05</em>
<em>The mean time is not greater than 60 minutes</em>
$9.40
They are paying 85% of the regular price..
so $7.99/ .85 = $9.40
