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choli [55]
3 years ago
14

Lament has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probab

ility that it will not be red?
Mathematics
2 answers:
umka2103 [35]3 years ago
8 0

Answer:

7/10 or 70%

Step-by-step explanation:

6 red chips, 10 blue chips, and 4 yellow chips = 20 chips

P( not red) = not red chips / total

                  = (blue + yellow) / total

                  = (10+4)/20

                  =14/20

                    =7/10

                   = 70%

tensa zangetsu [6.8K]3 years ago
8 0
Answer: 7/10

Explanation:
All chips together are 20/20 subtract 6 from the numerator and make 14/20 then divide to simplify giving you 7/10
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What is the equivalent of 1/50: fraction , in decimal form? We can determine the equivalent values of the fraction 1/50 in one simple way.
By dividing 1 and 50 to get the quotient of the digits given. <span><span>

1.  </span>1 / 50 = 0.02</span>

Therefore, 0.02 is the decimal form of the given fraction 1/50 and is a rational number.



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Vilka [71]

Distribute

6 × 2b + 6 × -4

Simplify 6 × 2b to 12b

12b + 6 × -4

Simplify 6 × -4 to -24

<u>12b - 24</u>

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3 years ago
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Calculate the distance between the points M(6,2) and N(-1,7).
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8 0
3 years ago
4x - Зу = -3<br> 4x + y = 17<br> Explain how as well
Jobisdone [24]

Step-by-step explanation:

I will solve your system by substitution.

(You can also solve this system by elimination.)

4x+y=17;4x−3y=−3

Step: Solve4x+y=17for y:

4x+y+−4x=17+−4x(Add -4x to both sides)

y=−4x+17

Step: Substitute−4x+17foryin4x−3y=−3:

4x−3y=−3

4x−3(−4x+17)=−3

16x−51=−3(Simplify both sides of the equation)

16x−51+51=−3+51(Add 51 to both sides)

16x=48

16/x = 48/16

(Divide both sides by 16)

x=3

Step: Substitute 3 for x in y=−4x+17:

y=−4x+17

y=(−4)(3)+17

y=5(Simplify both sides of the equation)

Answer:

x=3 and y=5

6 0
3 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
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