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3 years ago

Solve for x: -2x + 1 = -6

Mathematics

1 answer:

lukranit [14]3 years ago

8 0

<h2>Answer:</h2>

This is a 2-step equation. We will solve it in 2 steps.

$-2x + 1 = -6\\\\-2x = -7\\\\x = \frac{7}{2}$

The solution for this equation is x =<em> </em> $\frac{7}{2}$ .

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A psychologist designs an 8-item pain survey asking athletes to identify their level of pain after a football injury. She is

Maslowich

Answer:

<h3>There are $8!$ ways to answer.</h3>

Step-by-step explanation:

We know that the survey has 8 items, that means we can use the factorial number, because it's a simple permutation.

Using a factorial we have

$n!$ where $n=8$ , which is the total number of items.

$8! = 8\times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320$

Therefore, there are 40,320 ways to answer the survey, or 8! ways.

4 0

2 years ago

E) 340 people participated in a conference. If the ratio of ladies and gentleman is

mariarad [96]

Answer:

<em>There were 204 ladies and 136 gentlemen in the conference</em>

Step-by-step explanation:

Let's set the variables:

L = number of ladies in the conference

G = number of gentlemen in the conference

We know 340 people participated in the conference, thus:

L + G = 340 [1]

We are also given the ratio of ladies and gentleman as 3:2, thus:

L/G = 3/2

Cross-multiplying:

2L = 3G [2]

From [1]:

L = 340 - G

Substituting in [2]:

2(340 - G) = 3G

Operating:

680 - 2G = 3G

5G = 680

G = 680/5 = 136

G = 136

L = 340 - 136 = 204

L = 204

There were 204 ladies and 136 gentlemen in the conference

7 0

2 years ago

There are 58 students enrolled in an art class. The day before the class begins, 10.3% of the students cancel. How many students

Sidana [21]

First, set up a proportion. 10.3% is the percentage of students who cancelled, but the question is asking for the students who are attending, so let’s use 89.7% to make it easier. The proportion would be x over 58 is equal to 89.7% over 100%. Multiply 58 by 89.7, giving you 5,202.6. Next, take 5,202.6 and divide it by 100. This should give you 52. 026, but it’s best to round this to 52. 52 students actually attend the art class, making this your final answer.

7 0

3 years ago

A Lights-A-Lot quality inspector examines a sample of 25 strings of lights and finds that 6 are defective. What is the experimen

finlep [7]

Answer:

A 0.24 probability that one string is defective

Step-by-step explanation:

The sample probabiity:

$P=\frac{6}{25}$

$P=0.24$

If important to notice that this probability is from the sample, it can be extrapolated to the population but may not be the same.

5 0

3 years ago

Read 2 more answers

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:

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

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

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

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6 0

3 years ago