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3 years ago

6

Martin draws a card from a standard deck of cards, does not replace the card, and draws another card. What is the probability he

draws an ace card and a queen?

1 answer:

Zepler [3.9K]3 years ago

6 0

It is a 1/221 chance of drawing an ace and a queen

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Multiply 5.3 by 104.

Answer: 551.2

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On May 15, 2014, Jason made a water bill payment of $64.56. What is his balance after this transaction?

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3 years ago

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1. Simplify:<br>4(4y-7y^{2})-9(5y+2)<br><br>2. Simplify:<br>24 – 4(5y – 6z) + 3y – 7z

Greeley [361]

Answer:

1)-

How to solve your question

Your question is

4(4−72)−9(5+2)

4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)

Simplify

1

Rearrange terms

4(4−72)−9(5+2)

4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)

4(−72+4)−9(5+2)

4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)

2

Distribute

4(−72+4)−9(5+2)

{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)

−282+16−9(5+2)

{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)

3

Distribute

−282+16−9(5+2)

-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)

−282+16−45−18

-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18

4

Combine like terms

2)

−17y+17z+24

See steps

Step by Step Solution:

STEP1:Equation at the end of step 1

((24 - 4 • (5y - 6z)) + 3y) - 7z

STEP2:

Final result :

-17y + 17z + 24

−282+16−45−18

-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18

−282−29−18

-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18

Solution

−282−29−18

4 0

2 years ago

1-12 I don’t know if you can please help me thanks

katovenus [111]

1 24

2 .41

3 180

4 2000

5 46

6 .18

7 60.5666

8 6.5

9 128

10 .8

11 3.628739

12 4000

pretty sure juh looked it up hope it helps!

4 0

3 years ago

Match each spherical volume to the largest cross sectional area of that sphere

zlopas [31]

Answer:

Part 1) $324\pi\ units^{2}$ ------> $7,776\pi\ units^{3}$

Part 2) $36\pi\ units^{2}$ ------> $288\pi\ units^{3}$

Part 3) $81\pi\ units^{2}$ ------> $972\pi\ units^{3}$

Part 4) $144\pi\ units^{2}$ ------> $2,304\pi\ units^{3}$

Step-by-step explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

$A=324\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$324\pi=\pi r^{2}$

Solve for r

$r^{2}=324$

$r=18\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=18\ units$

substitute

$V=\frac{4}{3}\pi (18)^{3}$

$V=7,776\pi\ units^{3}$

Part 2) we have

$A=36\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$36\pi=\pi r^{2}$

Solve for r

$r^{2}=36$

$r=6\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=6\ units$

substitute

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ units^{3}$

Part 3) we have

$A=81\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$81\pi=\pi r^{2}$

Solve for r

$r^{2}=81$

$r=9\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=9\ units$

substitute

$V=\frac{4}{3}\pi (9)^{3}$

$V=972\pi\ units^{3}$

Part 4) we have

$A=144\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$144\pi=\pi r^{2}$

Solve for r

$r^{2}=144$

$r=12\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=12\ units$

substitute

$V=\frac{4}{3}\pi (12)^{3}$

$V=2,304\pi\ units^{3}$

5 0

3 years ago

Read 2 more answers