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3 years ago

Which equation can be solved using the expression

Mathematics

1 answer:

jenyasd209 [6]3 years ago

4 0

Wheres the other part of the question?

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Given the point and it''s image, determine the scale factor. A (3, 6) A'' (4.5, 9) G'' (3, 6) G (1.5, 3) B (2, 5) B'' (1, 2.5).

Natasha2012 [34]

Put the value of the letters with " over the original
So for A we have 3 in the front and 4.5 for A"
4.5/3 = 1.5
We can check if this is actually the scale factor but using the other numbers
The second numbers for A and A" are 6 and 9
6 * 1.5 = 9
So for point A it is 1.5
Use the same thing for G. That gives you 2
For B, you get 0.5
So the scale factors are 1.5, 2, and 0.5

5 0

3 years ago

Using the quadratic formula to solve 11x²-4x=1 What are The values of X?

Anvisha [2.4K]

$11x^2 - 4x = 1$
$11x^{2} - 4x - 1 = 1 - 1$
$11x^{2} - 4x - 1 = 0$
$x = \frac{-(-4) \± \sqrt{(-4)^{2} - 4(11)(-1)}}{2(11)}$
$x = \frac{4 \± \sqrt{16 + 44}}{22}$
$x = \frac{4 \± \sqrt{60}}{22}$
$x = \frac{4 \± 2\sqrt{15}}{22}$
$x = \frac{2 \± \sqrt{15}}{11}$
$x = \frac{2 + \sqrt{15}}{11}\ \bold{or}\ x = \frac{2 - \sqrt{15}}{11}$

8 0

3 years ago

Read 2 more answers

In 2012, a competition sponsored 301 sporting events. This was 256 more events than there were in an 1896 competition. How many

Vika [28.1K]

There were 54 sporting events in the year 1896 hosted by the competition

What is the function to determine the number of events in 1896?

The fact that events which are 301 sporting events in 2012 are 256 more than that of the events in 1896, means that the number of events in 1896 is 256 less than that of 2012

number of events in 1896=number of events in 2012-256

number of events in 2012=310

number of events in 1896=310-256

number of events in 1896=54 events

Find out more about more about number operations on:brainly.com/question/4344214

#SPJ1

4 0

1 year ago

Please helppppp!!!<br> Convert 2/7 to a decimal.<br> Round to the nearest hundredth.

horrorfan [7]

ANSWER:
0.29

STEP BY STEP:
Just divide 2/7 then round

you get 0.28571428571 when dividing
then round to 0.29

hope this helps

6 0

2 years ago

Read 2 more answers

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

Setler [38]

Answer:

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

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