A recent national survey found that parents read an average (mean) of 10 books per month to their children under five years old. The population standard deviation is 5. The distribution of books read per month follows the normal distribution. A random sample of 25 households revealed that the mean number of books read last month was 12. At the .01 significance level, can we conclude that parents read more than the average number of books to their children

Step-by-step explanation:

Null hypothesis: u = 10

Alternative hypothesis: u =/ 10

Using the formula: t = (x - u) / (s /√n)

Where x = 12, u = 10, s = 5 and n = 25

t= (12-10) / (5/√25)

t = (2)/(5/5)

t = 2/1= 2

t = 2.0

At a 0.01 level of significance with a degree of freedom of 24, the p-value is 0.0569, which is greater than 0.01 we will fail to reject the null and conclude that parents do not read more than the average number of books to their children

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct