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3 years ago

13

VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV this is middle school math

1 answer:

Yuri [45]3 years ago

5 0

Answer:

first method

<u>2</u><u>2</u>=<u>5</u>

7. x

we cross multiply to get

<u>2</u><u>2</u><u>x</u>= <u>5</u><u>×</u><u>7</u>

22. 22.

x= <u>3</u><u>5</u>

22

=1.59

second method

<u>2</u><u>2</u>=<u>5</u>

7. x

we multiply the denominators to get 7x

then we multiply each term by 7x

7x×<u>2</u><u>2</u> = <u>5</u>×7x

7. x

here the 7 and 7 will cancel out and the x and x will cancel out to get

<u>2</u><u>2</u><u>x</u>= <u>3</u><u>5</u>

22. 22

= 1.59

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

What is the solution to the equation 4x= 48?<br> Ox=12<br> OX= 10<br> O x= 44<br> Ox=8

svetoff [14.1K]

Answer:

12

Step-by-step explanation:

48/4=12 x=12

6 0

3 years ago

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3 years ago

Read 2 more answers

Prove the polynomial identity.

aleksandr82 [10.1K]

Explanation:

Since we have given that

$(2x+1)^2-2(2x+1)$

Now, we have to prove it as

$(2x+1)(2x-1)$

Now, start the iterations:

$(2x+1)^2-2(2x+1)\\\\=4x^2+1+4x-4x-2\\\\=4x^2+1-2\\\\=4x^2-1\\\\=(2x)^2-1\\\\=(2x+1)(2x-1)$

So, first blank will be filled with

$4x^2+4x+1$

Second blank will be filled with

$4x^2-1$

Hence proved.

7 0

3 years ago

Lee scored 91 points on a math testOn the next test, she scored 13 fewer pointsOn the third test, she scored 92What is the diffe

max2010maxim [7]

Answer:

14 points

Step-by-step explanation:

The second test Lee did was 13 points fewer than the first test, which was 91 points, so the second test's score is 91 - 13 = 78 points.

The third test had the score of 92, so this is the highest score, and the lowest score is the second one: 78 points.

The difference between Lee's highest and lowest scores is:

92 - 78 = 14 points

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3 years ago