How many solutions does the equation have? How do you know?
6x – 2 = 2(3x – 1)

infinite solutions

Step-by-step explanation:

First, solve the equation.

6x - 2 = 2(3x - 1)

6x - 2 = 6x - 2

By this point, no further work needs to be done because you have your answer. If these two equations are the exact same and are equal to each other, then there is an infinite number of solutions for x.

This is what'll happen if you continue to solve for x:

6x = 6x

x = x

This is infinite because if you substitute any real number, the equation is true.

For example: 1 = 1 or 176 + 59 = 176 + 59.

87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68.

91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80.

5.48% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

So, lets see each question:

What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?

The Institute reported that entry level wages for male college graduates were $21.68 per hour, with a standard deviation of 2.30.

So

This probability is the pvalue of Z when subtracted by the pvalue of Z when

X = 22.18

has a pvalue of 0.9382.

X = 21.18

has a pvalue of 0.0618.

There is a 0.9382 - 0.0618 = 0.8764 = 87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68.

What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?

This probability is the pvalue of subtracted by the pvalue of , following the same logic as the question above.

There is a 0.9573 - 0.0427 = 0.9146 = 91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80.

What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

This is the pvalue of Z when

So

has a pvalue of 0.0548.

So there is a 5.48% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.