Ask question

Ask question

All categories

4 years ago

12

A bag contains eight blue marbles six Green marbles 12 yellow marbles and 10 orange marbles a marble is drawn at random from the

bag what is the probability that the marble drone will not be blue

1 answer:

Lapatulllka [165]4 years ago

8 0

Answer:

120, or 1.2

Step-by-step explanation:

12 divided by 10 is 1.2 or do 12 times 10 which is 120

You might be interested in

Identify the area of the figure rounded to the nearest tenth. HELP?!! I don't understand!! Please help and show your work!!

Alex73 [517]

Answer:

$67.5\ cm^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle (figure 1) plus the area of trapezoid (figure 2)

see the attached figure to better understand the problem

The area of the rectangle is

$A=15*3=45\ cm^{2}$

The area of the trapezoid is

$A=\frac{1}{2}[(15-9)+3)](8-3)$

$A=\frac{1}{2}[6+3)](5)=22.5\ cm^{2}$

The area of the figure is

$45\ cm^{2}+22.5\ cm^{2}=67.5\ cm^{2}$

8 0

3 years ago

2830000000 who can write this number in “Scientific Notation.”

Genrish500 [490]

Answer:

$2.83*10^{9}$

Step-by-step explanation:

4 0

4 years ago

<img src="https://tex.z-dn.net/?f=5%5Cfrac%7B42%7D%7B170%7D%20%20-%203%5Cfrac%7B80%7D%7B170%7D%20" id="TexFormula1" title="5\fra

Reika [66]

$5\dfrac{42}{170}-3\dfrac{80}{170}=(5-3)+\left(\dfrac{42}{170}-\dfrac{80}{170}\right)\\\\=2+\left(-\dfrac{38}{170}\right)=2-\dfrac{32}{170}=1\dfrac{170}{170}-\dfrac{38}{170}\\\\=1\dfrac{170-38}{170}=1\dfrac{132}{170}=1\dfrac{132:2}{170:2}=1\dfrac{66}{85}$

$5\dfrac{42}{170}-3\dfrac{80}{170}=4\dfrac{170+42}{170}-3\dfrac{80}{170}\\\\=(4-3)+\left(\dfrac{212}{170}-\dfrac{80}{170}\right)=1+\dfrac{212-80}{170}\\\\=1+\dfrac{132}{170}=1\dfrac{132:2}{170:2}=1\dfrac{66}{85}$

7 0

3 years ago

An urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn (one by one) from the urn without replacement.

strojnjashka [21]

Answer: The required probability is $\dfrac{2}{15}.$

Step-by-step explanation: Given that an urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn one by one from the urn without replacement.

We are to find the probability that both drawn marbles are black.

Let E and F denote the events of two marbles one by one without replacement and let S and S' denote the corresponding sample spaces.

Then, we have

$n(E)=^4C_1=4,\\\\n(F)=^3C_1=3,\\\\n(S)=^{10}C_1=10,\\\\n(S')=^9C_1=9.$

Therefore, the probability that both marbles are red is given by

$p=P(E)\times P(F)=\dfrac{n(E)}{n(S)}\times\dfrac{n(F)}{n(S')}=\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{2}{15}.$

Thus, the required probability is $\dfrac{2}{15}.$

4 0

3 years ago

Pls ill give brainelist

goblinko [34]

Answer:

Vertical angle

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers