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3 years ago

PLEASE HELP WILL MAK BRAINLIEST

Mathematics

2 answers:

Darina [25.2K]3 years ago

8 0

Answer:

-5.6W+3.5

Step-by-step explanation:

$4+W-6.6W-1/2=\\4+W-6.6W-0.5=\\W-6.6W+4-0.5=\\-5.6W+3.5$

Basile [38]3 years ago

5 0

Answer: -5.6W+3.5

4+W-6.6W-0.5
=-5.6W+3.5

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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

Three more than the product of a number and 9 is equal to 7.

4vir4ik [10]

Answer:

(x*9)+3=7

Step-by-step explanation:

three more is +3

product of a number and 9 is x*9

and equal to seven is =7

8 0

3 years ago

Find the first term, a1, of an arithmetic sequence if a12 = 38 and a45 = 170.

snow_lady [41]

Standard formula for arithmetic sequence:
an = a0 + d(n-1)

if we use the two terms given, setting a12 as starting term and a45 as the end term an.

170 = 38 + 33d
170 - 38 = 33d
132 = 33d

132/33 = d
This is the common difference, use it to find the first term.

38 = a0 + (132/33)(12-1)
38 = a0 + (132/33)(11)
38 = a0 + 132/3
38 - 132/3 = a0
38 - 44 = a0
-6 = a0

The starting term is -6

8 0

3 years ago

4x²-y²=171 and 2x-y=9, value of 2x+y

11111nata11111 [884]

So you will need to solve for x and y before evaluating 2x+y....

2x-y=9, y=2x-9 now this will make 4x^2-y^2=171 become:

4x^2-(2x-9)^2=171

4x^2-(4x^2-36x+81)=171

36x-81=171

36x=252

x=7, now we can use 2x-y=9 to solve for y...

2(7)-y=9

14-y=9

-y=-5

y=5

now we know that x=7 and y=5, 2x+y becomes:

2(7)+5

14+5

19

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3 years ago

Greg bought 2 boxes of balloons. He used half of them to decorate his yard. he used 40 to decorate his porch. and used the rest

Anuta_ua [19.1K]

20 I'm guessing? Hope this helps!

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3 years ago