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Andre45 [30]
3 years ago
8

For what values of m does the equation mx = 5 have a single solution? can m have a value for which the equation does not have so

lutions; have infinite number of solutions?
Mathematics
1 answer:
Ratling [72]3 years ago
5 0

Answer:

Given equation: mx =5           ......[1]

To find for which values of m does the given equation have a single solution.

Divide equation [1] both sides by x,

\frac{mx}{x} = \frac{5}{x}

Simplify:

m = \frac{5}{x}

The value of m have a single solution for all real values of x except x≠0.

No solution states that there is no answer to the equation

Infinite solutions  states that any value for the variable would make the equation true.

we can write equation [1] as;

x= \frac{5}{m}

For the value of m = 0 , the given equation does not have solutions.

For any constant parameter m ,

x=\frac{5}{m}, the given equation does not have any infinite number of solutions.

   

 

   

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Natali5045456 [20]
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6 0
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In a soccer tournament teams receive 6 points for winning a game 3 points for trying a game and 1 point for each goal they score
Alex777 [14]
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8 0
2 years ago
Suppose that Upper X has a discrete uniform distribution f left-parenthesis x right-parenthesis equals StartLayout left-brace1st
emmasim [6.3K]

Answer:

the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

Step-by-step explanation:

Given the data in the question;

x        f(x)         xp(x)               x²p(x)

1         1/3        0.33333        0.33333

2        1/3        0.66667        1.33333

3        1/3        1.00000        3.0000

∑                    2.0000          4.6667

∑(xp(x)) = 2

∑(x²p(x)) = 4.6667

Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667

standard deviation σ = √variance = √0.6667 = 0.8165

Now since, n = 33 which is greater than 30, we can use normal approximation

for normal distribution z score ( x-μ)/σ

mean μ = 2

standard deviation = 0.817

sample size n = 33

standard of error σₓ = σ/√n = 0.817/√33 = 0.1422

so probability will be;

p( 2.1  < X < 2.4 ) = p(( 2.1-2)/0.1422) <  x"-μ/σₓ  <  p(( 2.4-2)/0.1422)

= 0.70 < Z < 2.81    

=  1 - ( 0.703 < Z < 2.812 )

FROM Z-SC0RE TABLE

=  1 - ( 0.25804 + 0.49752 )

= 1 - 0.75556

p( 2.1  < X < 2.4 ) = 0.2444

Therefore,  the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

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3 years ago
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katrin2010 [14]

Answer:

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Step-by-step explanation:

5 0
3 years ago
Match the information on the left with the appropriate equation on the right.
zaharov [31]

Answer:

See attachment below

Step-by-step explanation:

For the first option, we are given m = - 2 / 3, b = 3. This is likely in the point - slope form y = mx + b, where m = slope and b = y - intercept. Thus, the equation of the line in point - slope form given this information, should be y = - 2 / 3x + 3. It seems as if none of the following equations match this form, so let us interchange the equation a bit,

y = - 2 / 3x + 3,

y + 2 / 3x = 3,

2x + 3y = 9 - Option C

_______________________________________________________

Next we are given m = - 3 / 2, and that this line passes through the point ( 4, - 1 ). Substitute m as - 3 / 2, x as 4, y as - 1, knowing point ( 4, - 1 ), into the form y = mx + b - solving for b.

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_______________________________________________________

( 6, 3 ) and ( 3, 1 ) is given to lie on this line. The slope of the line should be as follows,

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Now let us determine the y - intercept ( b ) by substitute one of the points, say ( 6, 3 ). In this case x = 6, and y = 3,

3 = ( 2 / 3 )( 6 ) + b,

b = - 1,

y = 2 / 3x - 1 = Option D

Take a look at the attachment below for further help;

6 0
2 years ago
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