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BlackZzzverrR [31]
2 years ago
13

A bag contains 10 blue counting chips and 10 red counting chips. What is the probability that, without looking, Miguel will pick

a blue chip first, which he will keep, and on his second draw Miguel will select a red chip?
1/10


11/21


5/38


5/19
Mathematics
1 answer:
Aleksandr [31]2 years ago
5 0

Answer: 5/19

Step-by-step explanation:

(probability of blue chip)/(total chips) x (probability of red chip)/(new total chips)

10/20 x 10/19 = 100/380

simplify: 5/19

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denis23 [38]

Answer:

Angle WTS will be 180 degree minus 15x. You can write this as an expression:

180-15x

Now you have an expression for each angle in the triangle. Triangles will always have 180 degrees in them. You can add the expressions together and set it equal to 180. Then you can solve the equation for the x variable.

Step-by-step explanation:

Angle SWT + angle WTS + angle TSW = 180 degrees

Replace each angle with the expression then solve.

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27 bags is not the answer
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It's 27 bags your teacher just dum

Step-by-step explanation:

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On a test, you correctly answer six 5-point questions and eight 2-point questions. You earn 92% of the possible points on the te
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2 years ago
Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).
Serjik [45]

Answer:

\displaystyle f(x)=x^2+2x+2

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

\displaystyle f(x)=ax^2+bx+c

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

\displaystyle f(-1)=a(-1)^2+b(-1)+c

\displaystyle f(-1)=a-b+c

\displaystyle a-b+c=1.....[eq\ 1]

Now we use the second condition f(1)=5

\displaystyle f(1)=a(1)^2+b(1)+c

\displaystyle f(1)=a+b+c

\displaystyle a+b+c=5.......[eq\ 2]

Finally, we use the third condition f(2)=10

\displaystyle f(2)=a(2)^2+b(2)+c

\displaystyle f(2)=4a+2b+c

\displaystyle 4a+2b+c=10....[eq\ 3]

We put together eq 1, eq 2, and eq 3 to form the system

\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.

Adding the first two equations we have

\displaystyle 2a+2c=6

\displaystyle a+c=3

And also

\displaystyle b=2

Using the above equation and the value of b in the third equation, we have

\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.

Subtracting the first equation from the second

\displaystyle 3a=3

\displaystyle a=1

And therefore

\displaystyle c=2

Now we have all the values, the quadratic function is

\displaystyle \boxed{f(x)=x^2+2x+2}

6 0
3 years ago
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