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Alexeev081 [22]

3 years ago

12

What is the estimate for the probability of selecting a red crayon OR a yellow crayon from the bag?

1 answer:

pogonyaev3 years ago

7 0

Answer: It would be both because there is no number to it

Step-by-step explanation:

For example if there were 15 red crayons and 10 yellow ones it would be the red crayons because there is a bigger amount of them in the bag but because there is no number you have the same possibility of picking both.

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A bag contains 42 red marbles, 6 white marbles, and 8 gray marbles. You randomly pick out a marble, record its color, and put it

Elanso [62]

For any single draw,

$\mathbb P(\text{white})=\dfrac6{42+6+8}=\dfrac6{56}$
$\mathbb P(\text{gray})=\dfrac8{42+6+8}=\dfrac8{56}$

Drawing a white marble or a gray marble are disjoint events; only one of them can happen. So

$\mathbb P(\text{white or gray})=\mathbb P(\text{white})+\mathbb P(\text{gray})-\underbrace{\mathbb P(\text{white and gray})}_0$
$\mathbb P(\text{white or gray})=\dfrac6{56}+\dfrac8{56}=\dfrac{14}{56}$

Out of 224 draws, you should expect $\dfrac{14}{56}\times224=56$ of the marbles to be either white or gray.

6 0

4 years ago

Read 2 more answers

Solving for matrices

muminat

Answer:

D

Step-by-step explanation:

The augmented matrix for the system of three equaitons is

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)$

Multiply the first row by 5, the second row by -3 and add these two rows:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)$

Subtract the third row from the second:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)$

Divide the third row by 6:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)$

Now multiply the third equation by 26 and add it to the second row:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)$

You get the system of three equations:

$\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.$

From the third equation

$z=\dfrac{90}{45}=2.$

Substitute z=2 into the second equation:

$-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.$

Now substitute z=2 and y=5 into the first equation:

$3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.$

The solution is (1,5,2)

4 0

3 years ago

8th grade math Ik but I do need help though

kramer

Answer:

x =-4

Step-by-step explanation:

4x+9 = -2x-15

Add 2x to each side

4x+2x +9 = -2x+2x-15

6x+9 = -15

Subtract 9 from each side

6x +9-9=-15-9

6x = -24

Divide each side by 6

6x/6 = -24/6

x = -4

7 0

3 years ago

What is the y - intercept of the equation y = 8x+4?<br> x-1

Tomtit [17]

Answer:

4

Step-by-step explanation:

y=8x+4

y=mx+b where m=slope and b=y-intercept

b=4

3 0

4 years ago

1. Use the clues below to determine Luke's house number

Olegator [25]

I think it is b
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7 0

3 years ago

Read 2 more answers