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3 years ago

11

Joanna needs a book to prop up a table leg. She randomly selects a book , and selects another book . What is probability that sh

e will choose 2 math books? If there are 17 books and 8 are math

1 answer:

Stolb23 [73]3 years ago

3 0

Id say 40%
because 8 are math so 9 of them are other

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What is the smallest fraction out of 1/3, 1/6 and 2/9

Nastasia [14]

2/9 is the smallest because the denominator is the smaller meaning there are more places so in perspective its smallest

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3 years ago

The interior angles formed by the sides of a hexagon have measures that sum to 720 degrees. What is the measure of angle F?

aniked [119]

The sum of the angles shown is ...
720° = 3x + 240°
Then
480° = 3x
160° = x

Angle F = 160° -20° = 140°

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2 years ago

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Anyone, please help me please answer my question because I have to pass this tomorrow morning:(

Wittaler [7]

Step-by-step explanation:

If you need help with how I got my answer, you can ask me.

$\frac{2yx {}^{ - 4} }{(x {}^{ - 4}y {}^{4}) {}^{3} \times 2x {}^{ - 1}y {}^{ - 3} }$

$\frac{2yx {}^{ - 4} }{x {}^{ - 12}y {}^{12} \times 2x {}^{ - 1}y {}^{ - 3} }$

$\frac{2y x {}^{ - 4} }{2x {}^{ - 13} y {}^{ - 9} }$

$= x {}^{9} y {}^{ - 8}$

$= \frac{x {}^{9} }{y {}^{8} }$

12.

$( \frac{2u {}^{ 4} }{ - u {}^{2}v {}^{ - 1} \times 2uv {}^{ - 4} } ) {}^{ - 1}$

$( \frac{2u {}^{4} \times 1 }{ - 2 {u}^{3}v {}^{ - 5} } ) {}^{ - 1}$

$( - u v {}^{ 5} ) {}^{ - 1}$

$= - u {}^{ - 1} v {}^{ - 5} = - \frac{ 1}{uv {}^{5} }$

13.

$- \frac{2m {}^{4}n {}^{ - 1} }{( - m {}^{2}n {}^{ - 2}) {}^{ - 1} \times - nm {}^{ - 3} }$

$- \frac{2m {}^{4}n {}^{ - 1} }{ - m {}^{ - 2} n {}^{2} \times - nm {}^{ - 3} }$

$- \frac{2m {}^{4} n {}^{ - 1} }{m {}^{ - 5}n {}^{3} } = - 2m {}^{9} n {}^{ - 4}$

$= - \frac{2m {}^{9} }{n {}^{4} }$

14.

$( \frac{x {}^{3}y {}^{ - 4} }{ - {x}^{2} y {}^{ - 3} \times yx {}^{0} } ) {}^{3}$

$( \frac{x {}^{3}y {}^{ - 4} }{ - {x}^{ - 2}y {}^{ - 2} } ) {}^{3}$

$( - {x}^{5} y {}^{ - 2} ) {}^{3} = - x {}^{15} y {}^{ - 6}$

$- \frac{x {}^{15} }{ {y}^{6} }$

15.

$- \frac{yx {}^{4} \times - {y}^{3} z { }^{ - 4} }{(z {y}^{2} ) {}^{4} }$

$- \frac{ - y {}^{4} x {}^{4} z {}^{ - 4} }{ {z}^{4} y {}^{8} } = - y {}^{4} x {}^{4} z {}^{ - 8} = - \frac{(xy) {}^{4} }{ {z}^{8} }$

16.

$\frac{h {}^{7}j {k}^{4} }{4h {}^{4} } = \frac{1}{4} h {}^{3} = \frac{h {}^{3} jk {}^{4} }{4}$

6 0

2 years ago

From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three

Serga [27]

Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, $2+2+8 = 12$ , and there are no other possibilities, unless you consider the various permutations of the terms.

So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.

Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.

So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of $5 \times 5 \times 5 = 5^3 = 125$ possible triplets.

Once we know the "good" cases and the total number of possible cases, the probability is simply computed as

$P = \cfrac{\text{number of favourable cases}}{\text{number of all possible cases}} = \cfrac{3}{125}$

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3 years ago

5y^2+5−4−4y^2+5y^2−4x^3 −1

vredina [299]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{4 {x}^{3} + 6 {y}^{2} }}}}}$

Step-by-step explanation:

$\sf{5 {y }^{2} + 5 - 4 - 4 {y}^{2} + 5 {y}^{2} - 4 {x}^{3} - 1}$

Collect like terms

⇒ $\sf{ 4 {x}^{3} + 5 {y}^{2} - 4 {y}^{2} + 5 {y}^{2} + 5 - 4 - 1}$

⇒ $\sf{4 {x}^{3} + {y}^{2} + 5 {y}^{2} + 5 - 4 - 1}$

⇒ $\sf{4 {x}^{3} + 6 {y}^{2} + 5 - 4 - 1}$

Calculate

⇒ $\sf{4 {x}^{3} + 6 {y}^{2} + 1 - 1}$

⇒ $\sf{4 {x}^{3} + 6 {y}^{2} }$

Hope I helped!

Best regards! :D

6 0

2 years ago