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Ierofanga [76]
3 years ago
9

In a bag are 3 red, 9 blue, 2 yellow and 3 green marbles. If you draw out a marble at random, what the probability that you will

draw a blue marble? P(blue)
Question 2 options:

3/17


9/17


2/17


14/17
Mathematics
1 answer:
dmitriy555 [2]3 years ago
6 0

Answer:

9/17

Step-by-step explanation:

there are 17 marbles in total 9 of them are blue therefore, 9/17 is the correct answer

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Simplify:<br><br> {(-8)^-4 ÷ 2^-8}^2
Aleksandr-060686 [28]

Answer:

ok so first we have to do whats in the brackets then we have to do to exponits so first

(-0.00024414062divided by 0.00390625)^2

-0.06249999872^2

-0.00390624984

Hope This Helps!!!

8 0
2 years ago
Read 2 more answers
3+(-8) -<br> How to solve 3 plus
storchak [24]

Answer:

3-8=-5

-5=-5

true

Step-by-step explanation:

7 0
3 years ago
Ollie used 1/2 cup of vegetable oil to make brownies. She used another 1/3 of oil to make muffins. How much more oil did she use
disa [49]

Answer:

1/6 cups more

Step-by-step explanation:

1/2=3/6

1/3=2/6

3/6-2/6=1/6

8 0
3 years ago
Here goes another question: 4/5%9/10 and please if you can with procces
love history [14]
When you divide a fraction by another fraction, it is the same thing as multiplying that fraction by its reciprocal.

The reciprocal is the 'flipped' version of a fraction where the numerator becomes the denominator and the denominator becomes the numerator.

for example, the reciprocal of 9/10 is 10/9

Knowing this information 4/5÷9/10 is the same as 4/5×10/9

Now we can solve...

\frac{4}{5} * \frac{10}{9} = \frac{40}{45} = \frac{8}{9}

Answer= 8/9

7 0
4 years ago
Read 2 more answers
How do i solve this? (by factoring)<br> x(x+3)=x+8
AfilCa [17]

\sf x\left(x+3\right)=x+8\\\\\\\\x^2+3x=x+8\\\\\\\\\sf \boxed{\sf\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides}}\\\\\\\\\sf x^2+3x-8=x+8-8\\\\\\\\\sf \boxed{\sf \mathrm{Simplify}}\\\\\\\\\sf x^2+3x-8=x\\\\\\\\\boxed{\sf \mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}}\\\\\\\\\sf x^2+3x-8-x=x-x\\\\\\\\\boxed{\sf \mathrm{Simplify}}\\\\\\\\\sf x^2+2x-8=0\\\\\\\\\boxed{\sf factor~ x^2+2x-8=0}\\\\\\\\\sf \left(x-2\right)\left(x+4\right)=0

\sf \boxed{\sf \mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0}\\\\\\\\\sf x-2=0\quad \mathrm{or}\quad \:x+4=0\\\\\\\\\boxed{\sf \{x=2\}}\\\\\boxed{\sf \{x=-4 \} }

3 0
3 years ago
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