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3 years ago

8

Find the value of m!!!

1 answer:

faust18 [17]3 years ago

7 0

Answer:

$m=158$

Step-by-step explanation:

Angles on a straight line add up to 180 degrees.

Hence,

$(m-29)+51=180$

Remove brackets:

$m-29+51=180$

$m+22=180$

Subtract 22 from both sides:

$m+22-22=180-22$

$m=158$

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8n = -3m + 1; n = -2, 2, 4

bogdanovich [222]

I am assuming you want to solve for m in each case

8n = -3m + 1

8(-2) = -3m + 1

-16 = -3m + 1

-3m = -17

m = $\frac{17}{3}$

8(2) = -3m + 1

16 = -3m + 1

-3m = 15

m = -5

8(4) = -3m + 1

32 = -3m + 1

-3m = 31

m = $\frac{31}{3}$

8 0

3 years ago

Read 2 more answers

Find the sum 14 + (-9) + (-20)

wlad13 [49]

It would be -15 I think just calculate it to make sure xx

3 0

3 years ago

Jerry has an insurance policy with a premium of $150 per month. in june, he's in an accident and receives a bill with a

EastWind [94]

Answer:

he will have to pay $650 in the month of june

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1 year ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

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

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

Hey guys I need some help; I would really appreciate it :)

Masja [62]

Their answer is correct so you need explanation?

4 0

3 years ago