About ten percent of users do not close Windows properly. Suppose that Windows is installed in a public library that is used by

I’m not really certain, but I think C could be the answer; my apologies if that's inaccurate.

Response:

D. The sidelines are parallel because they are perpendicular to a common line.

Justification:

According to the perpendicular transversal theorem, when a line is perpendicular to one of two parallel lines, it is also perpendicular to the other line. Furthermore, the converse of the theorem states that if two lines are perpendicular to the same line, they must be parallel. Therefore, the sidelines are indeed parallel and also perpendicular to this single line.

The answer is C since 24-3=21, followed by 21/1.50=14 

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!

The solution is 20 days, found as follows:
Each of the three friends initially makes 4 bags of dough. After 10 days, each of these bags is split into 4 new bags:
3 friends × 4 bags = 12 bags, and 12 bags × 4 = 48 bags
Then in the following 10 days, 48 bags are again divided into 4 bags each:
48 × 4 = 192 bags
Adding the periods: 10 days + 10 days = 20 days
Therefore, 192 bags are created after 20 days.