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2 years ago

Pls help I’ll brainlest Complete the activities below

Mathematics

1 answer:

DanielleElmas [232]2 years ago

7 0

Answer:

Step-by-step explanation:

( $x_{1}$ , $y_{1}$ )

( $x_{2}$ , $y_{2}$ )

rate of change or slope m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ = $\frac{rise}{run}$

y = mx + b

"b" is y-intercept

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The sum of three numbers is 21. Three times the smallest is 3 less than the largest, while the sum of the largest and smallest i

Lynna [10]

3, 6, 12
I guessed and checked. So if 3 is the smallest number, I would, multiply it by 3, which is 9. 9 is 3 less than the largest number which would be 12. 12 + 3 is 15. To find the third number, subtract 21 and 15 which is 6

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2 years ago

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What is m∠KNL? Enter your answer in the box. ° A horizontal line segment M K intersects with line segment J L at their midpoint

inna [77]

t N.

In the figure shown below

Answer:

A horizontal line segment M K intersects with line segment J L at their midpoint N.

∠J N M =(5x+2)°

∠ LN M=3( x+ 14)°

So, ∠J N M + ∠ LN M =180°[ These two angles form linear pair.Angles forming linear pair are supplementary.]

⇒5 x+ 2+ 3 (x+ 14) =180 [ By Substitution]

⇒ 5 x+2 +3 x+42°= 180°

⇒ 8 x=180°-44°

⇒8 x= 136°

⇒x= 136°÷8

⇒x=17°

So, ∠J N M =5×17 +2=87°

∠ LN M= 3×(17 +14)=3×31=93

∠J N M =∠K N L [Vertically opposite angles]

∠K N L=87°

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3 years ago

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Solve the system of equations x - y = 12 and x + 7y = -20 by combining the equations.

iogann1982 [59]

Answer:

x=16, y=-4

Step-by-step explanation:

-1(x-y=12)

x+7y=-20

-x+y=-12

x+7y=-20

8y=-32

y=-4

x-4=12

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2 years ago

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A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of childr

Ne4ueva [31]

Answer:

a) 6

b) 4

c) 3

Step-by-step explanation:

Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian.

To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times.

Let ak be the number of trials performed until 2 males are born in experiment k. That is,

a1= number of trials performed until 2 males are born in experiment 1

a2= number of trials performed until 2 males are born in experiment 2

and so on.

If a1 + a2 + … + an = N

we would expect Np females.

Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born).

So we would expect 2n = N(1-p), or

N/n = 2/(1-p)

But N/n is the average number of trials per experiment, that is, the expectation.

We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female.

Here we have the probability of having a male is half as likely as females. So

1-p = p/2 hence p=2/3

The expected number of trials would be

2/(1-2/3) = 2/(1/3) =6

This means the couple would have 6 children: 4 females (the first 4 trials) and 2 males (the last 2 trials).

Here the probability of having a female = probability of having a male = 1/2

The expected number of trials would be

2/(1/2) = 4

This means the couple would have 4 children: 2 females (the first 2 trials) and 2 males (the last 2 trials).

Here, 1-p = 2p so p=1/3

The expected number of trials would be

2/(1-1/3) = 2/(2/3) = 6/2 =3

This means the couple would have 3 children: 1 female (the first trial) and 2 males (the last 2 trials).

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3 years ago

What is the equation of the line that passes through the points (-1, 7) and (2, 10) in Standard Form?

Usimov [2.4K]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-7}{2-(-1)}\implies \cfrac{3}{2+1}\implies \cfrac{3}{3}\implies 1$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}$

just to point something out, is none of the options, however -x + y = 8, is one, though improper.

3 0

2 years ago