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Pachacha [2.7K]

3 years ago

13

The total completion time is 25 days. task a is planned to take seven days. task b is planned to take five days. task c is plann

ed to take eight days. the tasks have a serial relationship. how much slack does the project have?

1 answer:

natita [175]3 years ago

7 0

The project currently has <span>Positive 5 days slack.

Times needed for task A = 7 days
Times needed for task B = 5 days
Times needed for task C = 8 days

Total predicted days = 20 days

Since the project is required to be completed in 25 days, the company has 25 days - 20 days = 5 days positive</span>

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Assume you have noted the following prices for books and the number of pages that each book contains.Book Pages (x) Price (y)A 5

tigry1 [53]

$a) $y=0.00991 x+1.042$b) $r^2=0.7503^2=0.563$\\C) $r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503$$

x: 500, 700, 750, 590 , 540, 650, 480

y: 7.00, 7.50 , 9.00, 6.5, 7.50 , 7.0, 4.50

We want to create a linear model like this :

$y=m x+b$

Where

$m=\frac{S_{x y}}{S_{x x}}$

And:

$\begin{aligned}& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right)}{n} \\& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}\end{aligned}$

With these we can find the sums:

$\begin{aligned}& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}=2595100-\frac{4210^2}{7}=63085.714 \\& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right) n}{=} 30095-\frac{4210 * 49}{7}=625\end{aligned}$

And the slope would be:

$m=\frac{625}{63085.714}=0.00991$

Now we can find the means for x and y like this:

$\begin{aligned}& \bar{x}=\frac{\sum x_i}{n}=\frac{4210}{7}=601.429 \\& \bar{y}=\frac{\sum y_i}{n}=\frac{49}{7}=7\end{aligned}$

And we can find the intercept using this:

$b=\bar{y}-m \bar{x}=7-(0.00991 * 601.429)=1.042$

And the line would be:

$y=0.00991 x+1.042$

Part b

The correlation coefficient is given by:

$r=\frac{n\left(\sum x y\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n \sum x^2-\left(\sum x\right)^2\right]\left[n \sum y^2-\left(\sum y\right)^2\right]}}$

For our case we have this:

$\begin{aligned}& \mathrm{n}=7 \sum x=4210, \sum y=49, \sum x y=30095, \sum x^2=2595100, \sum y^2=354 \\& r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503\end{aligned}$

The determination coefficient is given by:

$r^2=0.7503^2=0.563$

Part c

$r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503$

Learn more about regression line to visit this link

brainly.com/question/7656407

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1 year ago

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount S present at time t

liubo4ka [24]

Answer:

a) - r=5%: $S=$ 5,136.10$

- r=4%: $S=$ 4,885.61$

- r=3%: $S=$ 4,647.34$

b) - r=5%: t=14 years

- r=4%: t=17 years [/tex]

- r=3%: t=23 years [/tex]

c) The amount obtained is

- Compuonded quarterly: $5,191.83

- Compuonded continously: $5,200.71

The latter is always greater, since the more often it is capitalized, the greater the effect of compound interest and the greater the capital that ends up accumulating.

Explanation:

The rate of accumulation of money is

$dS/dt=rS$

To calculate the amount of money accumulted in a period, we have to rearrange and integrate:

$\int dS/S=\int rdt=r \int dt\\\\ln(S)=C*r*t\\\\S=C*e^{rt}$

When t=0, S=S₀ (the initial capital).

$S=S_0=Ce^{r*0}=Ce^0=C\\\\C=S_0$

Now we have the equation for the capital in function of time:

$S=S_0e^{rt}$

a) For an initial capital of $4000 and for a period of five years, the amount of capital accumulated for this interest rates is:

- r=5%: $S=4000e^{0.05*5}=4000*e^{0.25}= 5,136.10$

- r=4%: $S=4000e^{0.04*5}=4000*e^{0.20}= 4,885.61$

- r=3%: $S=4000e^{0.03*5}=4000*e^{0.15}= 4,647.34$

b) We can express this as

$S=S_0e^{rt}\\\\2S_0=S_0e^{rt}\\\\2=e^{rt}\\\\ln(2)=rt\\\\t=ln(2)/r$

- r=5%: $t=ln(2)/0.05=14$

- r=4%: $t=ln(2)/0.04=17$

- r=3%: $t=ln(2)/0.03= 23$

c) When the interest is compuonded quarterly, the anual period is divided by 4. In 5 years, there are 4*5=20 periods of capitalization. The annual rate r=0.0525 to calculate the interest is also divided by 4:

$S = 4000 (1+(1/4)(0.0525))^{5*4}=4000(1.013125)^{20}\\\\S=4000*1.297958= 5,191.83$

If compuonded continously, we have:

$S=S_0e^{rt}=4000*e^{0.0525*5}=4000*1.3= 5,200.71$

The amount obtained is

- Compuonded quarterly: $5,191.83

- Compuonded continously: $5,200.71

The latter is always greater, since the more often it is capitalized, the greater the effect of compound interest and the greater the capital that ends up accumulating.

5 0

4 years ago

The financial budgets include the Select one: a. cash budget and the selling and administrative expense budget. b. cash budget a

Mekhanik [1.2K]

Answer:

The correct answer is B. Cash Budget and Budgeted Balance Sheet

Explanation:

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It includes the analysis of the investment, the projection of income and expenses and the form of financing. In this work, people who wish to start a business should keep in mind that this will be their greatest source of information.

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4 years ago