Let x denote the average number of hours per week that a student at Happy High School spends doing homework. Suppose the frequency distribution of x is normal (i. E., symmetrical and bell-shaped) with a mean of 3.4 hrs. And a standard deviation of 2.6 hrs. What percentage of the students at Happy High spend an average of less than 6 hours per week doing homework?

0.84134

Step-by-step explanation:

Given that:

Mean = 3.4 hours

Standard deviation (s) = 2.6 hours

Percentage of students that spend less than 6 hours doing homework

P(x < 6)

Using the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (6 - 3.4) / 2.6

Z = 2.6 / 2.6

Z = 1

p(Z < 1) = 0.84134 ( Z probability calculator)

The percentage of the students at Happy High spends an average of less than 6 hours per week doing homework is 0.84134

a

step-by-step explanation:

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x