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2 years ago

I will make u brainlist ? if u talk with me....

Mathematics

2 answers:

earnstyle [38]2 years ago

4 0

Answer:

hard

Step-by-step explanation:

I have never seen this before

zavuch27 [327]2 years ago

3 0

Answer:

Hello There

....

......... How you doing?

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M is the midpoint of QR<br><br> M = (-2,6)<br> Q= (8,-10)<br> R = ????

Zolol [24]

Answer:

R = (-12, 22)

Step-by-step explanation:

Unfortunately it's not a straight line, so gonna need to put in some extra work. Basically think of it like breaking it into it's horizontal and vertical components then doubling them.

From Q to M you move 10 spaces to the left, so to go from M to R you will go another 10 to the left. Similarly, you start at -10 for the y value and go up 16 to 6, and you will need to go up another 16 to get to R.

So for x, 8 -10 = -2 then -2 - 10 = -12, so R's x value is -12

y we do the same thing. Gonna do it in one step though. -10 + 16 + 16 = 22

So R is (-12, 22)

4 0

3 years ago

Two pipes running together can fill a tank in 20 minutes.

Hunter-Best [27]

Answer:

45, 36

Step-by-step explanation:

let x y minutes be the time for pipes to fill the tank, let n be the water needed to fill the tank.

x-y=9

(n/x)*20+(n/y)*20=n

n is removed by dividing the 2nd equation by n

here u get:

(1/x+1/y)*20=1

you sub x=9+y into the above equation

1/y+1/(9+y)=1/20

((9+y)+(y))*20=(9+y)(y)

180+40y=9y+y2

y2-31y-180=0

then use the quadratic formula and u will find y=36 or -5

-5 is rejected because it is negative

x=36+9=45

therefore it is 36 mins and 45 mins

3 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$