Suppose that only 65% of all drivers in a certain state wear a seat belt. a random sample of 80 drivers is selected. what is the probability that more than forty-two drivers wear a seat belt?

Pr(X >42) = Pr( Z > -2.344)

                = Pr( Z< 2.344) = 0.9905

Step-by-step explanation:

The scenario presented can be modeled by a binomial model;

The probability of success is, p = 0.65

There are n = 80 independent trials

Let X denote the number of drivers that wear a seat belt, then we are to find the probability that X is greater than 42;

Pr(X > 42)

In this case we can use the normal approximation to the binomial model;

mu = n*p = 80(0.65) = 52

sigma^2 = n*p*(1-p) = 18.2

Pr(X >42) = Pr( Z > -2.344)

                = Pr( Z< 2.344) = 0.9905

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.

answer: 33 ( approx)

step-by-step explanation:

since, in the linear regression formula, the slope is the a in the equation y = b + ax ( by comparing it with the general equation of line y = mx + c )

where,

here given table is,

price ($), x   :           18         23       25       29       30       33

pairs sold, y :         16         13         10       8         6       2

so, ,   ,   ,

and   ,

thus, slop of the given line is,

⇒  

⇒   ≈ 33