A car dealer investigate the association between the number of TV commercials he ran each week and the number of cars he sold the following weekend. He found the correlation to be r = 0.56. During this time he collected the data he ran an average of 12.4 commericals a week with a standard deviation of 1.8, and sold an average of 30.5 cars with a standard deviation of 4.2. Next weekend he is planning a sale, hoping to sell at least 40 cars. Create a linear model to estimate the number of commercials he should run this week.

We can boldly say, if the car dealer is hoping to sell 40 cars, the model predict he should run about 15 commercial this week.

Step-by-step explanation:

To start with solving this question, we try to define the parameters in the question:

So we have:

X = ? numbers of cars sold

Y = ? of commercial

X1 = 30.5

SX = 4.2

Y1 = 12.4

SY = 1.8

r = 0.56

Now, recalling the equation for slope and intercept

If we apply them in this instance,

we have:

Foe Slope b1 = r . SY / SX   and         For intercept = b₀ = Y1 - b , X1

           b1 = 0.56 (1.8) / 4.2                           b₀ = 12.4 - (0.24)(30.5)

            b1 = 0.24                                                 b₀ = 5.08

Where:

       Y¹ = b₀ + b1 X

        Y¹ = 5.08 + 0.24 X

Therefore, ? number of commercial = 5.08 + 0.24 cars

where:

? number of cars = 40

Therefore,

5.08 + 0.24 (40)

14.68

Hence, we can boldly say, if the car dealer is hoping to sell 40 cars, the model predict he should run about 15 commercial this week.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)