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Nata [24]
2 years ago
15

In a second grade class containing 13 girls and 9 boys, 2 students are selected at random to give out the math papers. What is t

he probability that the second student chosen is a boy, given that the first one was a girl?
Mathematics
2 answers:
Finger [1]2 years ago
8 0
13/22 x 9/21 = 0.25                                               
Rashid [163]2 years ago
4 0

Answer:

42.86% probability that the second student chosen is a boy, given that the first one was a girl.

Step-by-step explanation:

Initially, there are 22 students, of which 13 are girls and 9 are boys.

The first student chosen was a girl, so now there are 21 students, of which 12 are girls and 9 are boys.

What is the probability that the second student chosen is a boy, given that the first one was a girl?

There are 21 students, of which 9 are boys. So there is a 3/7 = 0.4286 = 42.86% probability that the second student chosen is a boy, given that the first one was a girl.

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. <br><br><br> Solve the equation. <br><br> 2/3 x – 6 = 9
Allisa [31]
Hey there!

Okay let's get cracking :D

The equation is....  (2x/3) - 6 = 9

2x/3 = 15
2x = 45
<u>x = 22.5</u><u />    <----- And that is your answer!

I hope it helped! 

<u>Please vote for me as Brainliest if there is a second answer!</u>
4 0
3 years ago
Read 2 more answers
Triangle PQR has vertices P(–2, 6), Q(–8, 4), and R(1, –2). It is translated according to the rule (x, y) → (x – 2, y – 16).
Luden [163]
The y-value is the one on the right. Therefore, the y-value of P is 6. Following the rule, the y-value of P' is 6-16= -10.
8 0
3 years ago
Read 2 more answers
Suppose we want to choose 5 letters, without replacement, from 9 distinct letters. (If necessary, consult a list of formulas.) (
lara [203]

Answer:

A.This is the number of combinations of 6 from 15

= 15C6

=  15! / (15-6)! 6!

= 5,005 ways.

B.  This is the number of permutaions of 6 from 15:

= 15! / (15-6)!

= 3,603,600 ways.

Step-by-step explanation:

This was someone else's work not mine sorry here's creditssss :))

brainly.com/question/15145413

5 0
3 years ago
D.
Alex

Answer:

Step-by-step explanation:

This is your correct answer if ur aim is simplification!

3 0
2 years ago
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS
Elina [12.6K]

You want to end up with A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:

A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with 2\sin(4\pi t)+5\cos(4\pi t). Right away, we know \omega=4\pi.

We also need to have

\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that \sin^2x+\cos^2x=1 for all x; this means

(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}

Then

\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)

So we end up with

2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)

8 0
2 years ago
Read 2 more answers
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