Answer:
Equal annual contributions to the college savings account over the next 18 years is : $4,745.6
Explanation:
Suppose the time the child was born is the Beginning of Year 0 (Y0). So, 18 equal contributions need to be made at the beginning of each year from Y0 to Year 17. Denote these cash flow as Annuity 1 which equal: ( C/ 2%) x ( 1.02^18 -1) = 21.4123 x C with C is the equal annual contribution
The tuition fee starting from the beginning of Year 18 end at the Beginning of Year 21 is a growing annuity at 2.5% growth rate. The Value of this annuity ( Annuity 2) discounted to the Beginning of Year 17 calculated as followed:
(28,000 / (2% - 2.5% ) x ( 1 - [( 1+2.5%)/(1+2%)]^4 ) = $110,614
To save enough for college fee, The future value of Annuity 2 must equal the present value of Annuity 2 calculated above.
Thus, we have: 21.4123 x C = 110,614 <=> C = $4,745.6
Answer:
B) higher, because more games are televised today.
Explanation:
Opportunity costs are the cost of choosing one alternative from another.
In this case, when college students attend college football games they are unable to do other activities while they are at the stadium or going to the stadium. The cost of those alternatives that are lost are higher now because many college football games are televised. So a student is now able to watch the game while doing other activities.
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Researchers in the health and social sciences can obtain their data by getting it directly from the subjects they’re interested in. This data they collect is called primary data. Another type of data that may help researchers is the data that has already been gathered by someone else. This is called secondary data. Hope this helped!
Explanation:
To find the probability that the sample which fails to meet the required weight or the standard weight of the marshmallows having banana flavor if the process of production is working, such that probability for the weekly sample leads to shutdown of the production if the process of the production is running properly of 1 % of the probability that at least five boxes out of the twenty five sample fails to meet the standard weight which is less than one percent that is 
We know that for p = 0.8, 
Now using binomial simulation, we can determine that
for p = 0.0452
So the production process is to be redesigned for reducing the percentage of boxes of the Go Bananas of 16 ounces which failed to meet the required weight of the marshmallows having banana flavor if the production process is working properly to 5.42 percent.