what would be the amount of a check issued for an 08/11 invoice for 648.50 paid on 09/13 with the terms: 3/10, 2/15, N30-5%APR?

Answer:

Let's explore this step-by-step:

Greetings!

Here are the variables

X: daily cost of hotel rooms

Y: expenses for entertainment

Refer to the second attachment for the scatter plot.

The population regression equation can be expressed as E(Yi)= α + βXi

To find the y-intercept and the slope of the regression line, apply the following formulas:

a= Y[bar]-bX[bar]

In this scenario, n= 9; ∑X= 945; ∑X²= 103325; ∑Y= 1134; ∑Y²= 148804; ∑XY= 123307

Calculating X[bar]= ∑X/n= 945/9= 105

Calculating Y[bar]= ∑Y/n= 1134/9= 126

Thus, it follows that a= 126 - 1.03*105= 17.49

The regression equation is then given by ^Y= 17.49 + 1.03Xi

Slope interpretation: For each unit increase in the daily hotel rate, the expected average spending on entertainment rises by $1.03.

Given that the room rate in Chicago is $128 (X), we can predict the entertainment expenditure (Y) using the estimated regression line:

^Y= 17.49 + 1.03Xi= 17.49 + 1.03*128= 149.33

The anticipated spending on entertainment for Chicago is $149.33

I trust this will assist you!

It appears you've gathered most of the relevant information, but a critical element is missing. The surface represented is parameterized by... and has a normal vector corresponding to this representation with a specified norm. Thus, the integral of... is...

<span>I'm fairly certain it's C $0.12


Good luck! I hope I was able to assist:)
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The equations 13x + 15y = 55.50 and 46x + 16y = 131.50 can be utilized to find out the cost per pound for both bananas and grapes, where x represents the price for bananas and y denotes the price for grapes.

c) Step-by-step breakdown: The collision rate is 1.2 incidents per 4 months, which can be expressed as 0.3 incidents monthly. Therefore, the Poisson distribution for the variable X representing monthly collisions is defined as P(X = x) =... for x ∈ N ∪ {0} = 0 otherwise. (1) Where X = 0 denotes no collisions during a 4-month timeframe, substituting gives P(X = 0) =... (2). For a 4-month period, P(No collision in 4 month period) =... (3). Two collisions in a 2-month span translate to 1 per month, thus P(X =1) =... (4). Over 2 months, P(2 collisions in a 2 month period) =... (5). One collision over a 6-month period equates to P(1 collision in 6 months period) =... (6). Consequently, P(1 collision in 6 month period) results in... (7). For no collisions in a 6-month period, P(No collision in 6 months period) =... (8). Finally, the probability of 1 or fewer collisions over six months is P(1 or fewer collision in 6 months period) = (8) + (7) = 0.0785 + 0.1653.