Which inequality has a solution that is the set of all real numbers?

Find the LCD to each the fractions and then add them together 1/4 + 2/5 + 1/3

jekas [21]

The LCD of these fractions is 60!

The sum of these fractions is $\frac{59}{60}$ !

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6 0

3 years ago

Read 2 more answers

An airline's public relations department claims that when luggage is lost, 88% is recovered and delivered to its owner within 24

melisa1 [442]

Answer:

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

Step-by-step explanation:

Test if the proportion of times that luggage is returned within 24 hours is less than 0. 88

At the null hypothesis, we test if the proportion is of 0.88, that is:

$H_0: p = 0.88$

At the alternate hypothesis, we test if this proportion is less than 0.88, that is:

$H_a: p < 0.88$

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.88 is tested at the null hypothesis:

This means that $\mu = 0.88, \sigma = \sqrt{0.88*0.12}$

A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day.

This means that $n = 160, X = \frac{138}{160} = 0.8625$

Test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.8625 - 0.88}{\frac{\sqrt{0.88*0.12}}{\sqrt{160}}}$

$z = -0.68$

P-value of the test:

The p-value of the test is the probability of finding a sample proportion below 0.8625, which is the p-value of z = -0.68.

Looking at the z-table, the p-value of z = -0.68 is of 0.2483.

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

7 0

2 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$