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Rama09 [41]
3 years ago
9

Classify the following as discrete or continuous random variables.

Mathematics
1 answer:
Elena-2011 [213]3 years ago
8 0

Answer:

A discrete variable is a variable that can only take some selected given values from a given set, like x ∈ {x₁, x₂, ..., xₙ}

While a continuous variable can take any value in the range, like x ∈ {x₁, xₙ}.

1)  The weight of bags of apples, with 10 apples in each bag.

Here the weight depends on the number of apples in the bag, and we can have only whole numbers of apples in the bag, so this is clearly a discrete variable.

2) The number of times required for a modem to dial an internet provider before connecting.

Same as before, here we can have only a whole number of attempts, so this set is discrete.

3) Out of 10 times connecting to an internet provider, the average number of attempts necessary before connecting.

Here we will have an average of 10 integer numbers:

M = \frac{x_1 + x_2 + ... + x_{10}}{10}

so here we have a kinda dense set of possible values, so we can think of this as a continuous variable.

4)  A pair of dice is rolled, and the sum to appear on the dice is recorded.

Here the possible values are whole numbers again, so we can not have values like 3.5, so again, this is a discrete variable.

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PLEASE HELP ME!!!!!!!!!! MUST SHOW ALL WORK
vekshin1
Do it by splitting middle term.
Then you will be able to get factors . You can learn how to split middle term by searching videos on you tube or any site.
7 0
3 years ago
Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

It is a homogeneous solution:

y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
1)35% of _____is 42.<br> 2)92% of_____ is 115.
dangina [55]
35% of 120 is 42. (42/32 then multiply that by 100)

92% of 125 is 115. (115/92 then multiply that by 100)
5 0
3 years ago
Read 2 more answers
Please help!!!<br> 20 points
mrs_skeptik [129]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
-3x - y = 15<br> y = -8x<br> (x, y) = (0<br> )<br> 9,0
Nataly [62]
<h3>The solution is (x, y) = (3, -24)</h3>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

-3x - y = 15 -------- eqn 1

y = -8x ------ eqn 2

We have to find solution of (x, y)

We can solve by substitution method

<em><u>Substitute eqn 2 in eqn 1</u></em>

-3x - (-8x) = 15

-3x + 8x = 15

5x = 15

Divide both sides by 5

<h3>x = 3</h3>

Substitute x = 3 in eqn 2

y = -8(3)

<h3>y = -24</h3>

Thus solution is (x, y) = (3, -24)

4 0
3 years ago
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