To determine the cost to fertilize the trees, you need to figure out how many kilograms of fertilizer are needed. With the information given, you can determine The number of grams needed to fertilize 1200 trees. To do this you would multiply 200 times 1200. This equals 240,000 grams. To convert this to kilograms, divide 240000g by 1000 g. Every group of 1000 g is 1 kg. The answer is 240 kilograms. Multiply 240 kg by the price of $2.75 per kilogram to get $660 as the cost.
Answer:
<h2>C. <em>
20,160</em></h2>
Step-by-step explanation:
This question bothers on permutation since we are to select a some people out of a group of people and then arrange in a straight line. If r object are to be arranged in a straight line when selecting them from n pool of objects. This can be done in nPr number of ways.
nPr = n!/(n-r)!
Selection of 6 people out of 8 people can therefore be done in 8C6 number of ways.
8P6 = 8!/(8-6)!
8P6 = 8!/2!
8P6 = 8*7*6*5*4*3*2!/2!
8P6 = 8*7*6*5*4*3
8P6 = 56*360
8P6 = 20,160
<em>Hence this can be done in 20,160 number of ways</em>
The outlier is 71.
An outlier in a set of data is the one that is significantly lower or higher than the average of the totals. Whichever number skews the average is the outlier.
In this case, it's 71.
:)
The capital formation of the investment function over a given period is the
accumulated capital for the period.
- (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.
- (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.
Reasons:
(a) The given investment function is presented as follows;
(a) The capital formation is given as follows;
From the end of the second year to the end of the fifth year, we have;
The end of the second year can be taken as the beginning of the third year.
Therefore, for the three years; Year 3, year 4, and year 5, we have;
The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87
(b) When the capital stock exceeds $100,000, we have;
![\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cmathbf%7B%5Cleft%5B1000%20%5Ccdot%20%20e%5E%7B0.1%20%5Ccdot%20t%7D%7D%20%2B%20C%20%5Cright%5D%5Et_0%7D%20%3D%20100%2C000)
Which gives;
The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.
Learn more investment function here:
brainly.com/question/25300925
This is one example of a trinomial with a leading coefficient of 3 and a constant term of -5
