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Sergio039 [100]
2 years ago
15

A graph titled Monthly Sales and Advertising Costs has Advertising Costs (1,000 dollars) on the x-axis and sales (1,000 dollars)

on the y-axis. A line goes through points (6.4, 117) and (6.6, 120).
The scatterplot and trend line show a positive correlation between advertising costs and sales.
Using the points (6.4, 117) and (6.6, 120), what is the slope of the trend line?
–15
–12
12
15
Mathematics
2 answers:
daser333 [38]2 years ago
8 1

AnswBrainly User

When x = 6.4, y=117, therefore

15*6.4 + b = 117

b = 117 -  96 = 21

Confirm this value by checking (6.6, 120).

15*6.6 + 21 = 120 (Correct)

Answer: b = 21

Step-by-step explanation:

Brianna bts
2 years ago
wrong
kenny6666 [7]2 years ago
7 0

Answer:

The real answer is 15.

Step-by-step explanation:

The kid above me thought they were smart using the old copy and paste method, and they did it from a completely different question. As you can tell, 21 isn't even an option, so 15 is the answer.

Brianna bts
2 years ago
correct tthank you
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Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

             = 8(1)

             = 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =  

              =  

              =  

              = 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

                = 1 - {+++++}

                = 1 - (){+++++}

                = 1 - (0.000335){+++++}

                = 1 - (0.000335){1+8+32+85.34+170.67+273.07}

                = 1 - (0.000335){570.08}

                = 1 - 0.1909

                = 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

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