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2 years ago

3(x + 4) = 27 what is x

Mathematics

1 answer:

Alla [95]2 years ago

6 0

Answer:

x = 5

Step-by-step explanation:

This is a simple equation if you look at it right and don't get scared by the "x."

All we need to do is figure out what times 3 = 27. We can do this easily by diving 27 by 3.

27 ÷ 3 = 9

Since our "equation" is 3(x + 4) = 27 we know whatever is in the parenthesis () has to equal to 9. We already get 4 so:

5 + 4 = 9.

3 times 9 = 27.

5 is your answer.

Hope this helps and have a nice day!

PS if you could mark brainliest that'd be great! I'm very close the "expert" rank. Haha, sorry I usually don't ask.

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Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanag

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Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4:The value of E(y) is 4.6667.

Part 5:The value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6:The value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as $f_{xy}(x,y)\neq f_x(x).f_y(y)\\$

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

$f(x,y)=\frac{1}{16}$ for $2\leq y\leq 2x\leq 10$

Now the distribution of x is given as

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy$

Here the limits for y are $2\leq y\leq 2x$ So the equation becomes

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5$

Part 2:

The probability is given as

$P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5$

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx$

Here the limits for x are $y/2\leq x\leq 5$ So the equation becomes

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10$

So the value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4

The value is given as

$E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667$

So the value of E(y) is 4.6667.

Part 5

This is given as

$f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}$

So the value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6

The value is given as

$\geq M_{x,y}(y)=E(f_{xy}(x))=\int\limits^5_{y/2} {x f_{xy}(x)} \, dx \\M_{x,y}(y)=\int\limits^5_{y/2} {x \frac{2}{10-y}} \, dx \\M_{x,y}(y)=\frac{2}{10-y}\left[\frac{x^2}{2}\right]^5_{\frac{y}{2}}\\M_{x,y}(y)=\frac{2}{10-y}\left(\frac{25}{2}-\frac{y^2}{8}\right)\\M_{x,y}(y)=\frac{y+10}{4}$

So the value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7

The value is given as

$E(x)=\int\limits^{5}_1 {xf_x(x)} \, dx\\E(x)=\int\limits^{5}_1 {x\frac{x-1}{8}} \, dx\\E(x)=\frac{1}{8}\left(\frac{124}{3}-12\right)\\E(x)=\frac{11}{3} =3.6667$

So the value of E(x) is 3.6667.

Part 8

The value is given as

$E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xyf_{x,y}(x,y)} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xy\frac{1}{16}} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \frac{x}{16}\left[\frac{y^2}{2}\right]^{10}_2\, dx\\E(x,y)=\int\limits^{5}_1 3x\, dx\\\\E(x,y)=3\left[\frac{x^2}{2}\right]^5_1\\E(x,y)=36$