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2 years ago

X + y = 3 . If y = 3, find the value of x

Mathematics

2 answers:

agasfer [191]2 years ago

6 0

Answer:

Step-by-step explanation:

x + y = 3

y = 3

=> x + 3 = 3

=> x = 3 - 3

=> x = 0

Aleks [24]2 years ago

6 0

I think the answers 0

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A system has one each of two different types of components in joint operation. Let X and Y denote the random lengths of the live

RSB [31]

Answer:

The value of E(Y/X) is 2.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

$f(x,y)=\left \{ {{\dfrac{x}{8}e^{-\dfrac{x+y}{2}} \,\,\,\,0 \leq x,0\leq y \atop {0}} \right.$

So the marginal function for X is given as

$f_x=\int\limits^{\infty}_0 {f(x,y)} \, dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x+y}{2} }dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x}{2}}e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{8}e^{-\frac{x}{2}}\int\limits^{\infty}_0e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{4}e^{-\frac{x}{2}}$

Now

$f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }$

Now the value of E(Y/X) is given as

$E(Y/X)=\int\limits^{\infty}_0 {yf_{Y/X}} \, dy \\E(Y/X)=\int\limits^{\infty}_0 {y\dfrac{1}{2}e^{-\frac{y}{2} } }\, dy\\E(Y/X)=\dfrac{1}{2}\dfrac{\sqrt{2}}{(\dfrac{1}{2})^2}=2$

So the value of E(Y/X) is 2.

3 0

2 years ago

What is wrong in the drawing? How can you make this work Correct?

Kryger [21]

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4 0

3 years ago

I need it ungently it’s due in 20 mins

Ipatiy [6.2K]

The answer is c you must look at key words i learned my key words

5 0

3 years ago

Question:

AleksandrR [38]

Answer:

x=8 and y=3 (So, yes!)

Step-by-step explanation:

I will solve your system by substitution.

(You can also solve this system by elimination.)

−x+4y=4;−x+3y=1

Step: Solve −x+4y=4 for x:

−x+4y+−4y=4+−4y(Add -4y to both sides)

−x=−4y+4

-x/-1 = -4y+4/-1 (Divide both sides by -1)

x=4y−4

Step: Substitute 4y−4 for x in −x+3y=1:

−x+3y=1

−(4y−4)+3y=1

−y+4=1(Simplify both sides of the equation)

−y+4+−4=1+−4(Add -4 to both sides)

−y=−3

-y/-1 = -3/-1 (Divide both sides by -1)

y=3

Step: Substitute 3 for y in x=4y−4:

=4y−4

x=(4)(3)−4

x=8(Simplify both sides of the equation)

Answer:

x=8 and y=3

4 0

1 year ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

2 years ago