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1 year ago

7

A deck of cards contains red cards numbered 1,2,3, Blue cards numbered 1,2,3,4,5,6, and green cards numbered 1,2. If a single ca

rd is picked at random, what is the probability that the card is not green? Provide the final answer as a fraction?

1 answer:

serg [7]1 year ago

4 0

Answer:

$\frac{9}{11}$

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We have:

3 red cards

6 blue cards

2 green cards.

What is the probability that the card is not green?

3 + 6 + 2 = 11 cards, of which 3 + 6 = 9 are not green. So

$\frac{9}{11}$ probability that the card is not green

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Answer:

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Step-by-step explanation:

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60 - 45 = 15 degrees

Since 1 hour equals an increase of 5 degrees we need to divide 15 by 5 to calculate how many hours before the temperature increases to 60 degrees

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2 years ago

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Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago

18÷90.54 I need help I have I minute!!!

Tcecarenko [31]

0.19880716 should be your answer
Sorry if wrong

5 0

2 years ago

Read 2 more answers