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3 years ago

Kiran scored 223 more points in a computer game than Tyler. If Kiran scored 409

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

186

Step-by-step explanation:

Tyler: 223 less than Kiran

Kiran: 409

409 - 223 = 186

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2(i<3u) is the answer to your problem

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4 years ago

Suppose the population of a town is 3,400 in 2000. The population decreases at a rate of 2% every 20 years. What will be the pro

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The formula would be
A=p (1-r)^t

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3 years ago

LifeCharge Batteries claims that their batteries last an average of 210 hours when used in a hand held gaming system, with a sta

Irina-Kira [14]

Answer:

The answer is "0.0023 and Choice a"

Step-by-step explanation:

In point a:

The central limit theorem:

$\mu_{X}= 250\\\\\sigma_{X}=\frac{\sigma}{\sqrt{n}}=\frac{6}{\sqrt{100}} =0.6\\\\P(\bar{X}>251.7)\\\\=P(z>\frac{251.70-250}{0.6})\\\\=P(z>2.83)\\\\=1-0.9977\\\\=0.0023$

In point b:

The chance is lower than 0.05. This means that it is very rare to see the average sample above 251.8 if the argument is valid, that is why the means of the sample is far away from the common claimed and casts doubt on the actual mean.

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3 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:

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

- 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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3 years ago

1/2y=4(x-1/4) express y in terms of x

Sloan [31]

I have know clue what the answer to this question is

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4 years ago