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True [87]
3 years ago
5

Given a bag of 20 marbles containing 8 green, 4 red, 2 blue, and 6 yellow, of a person picks out a single marble from the bag wi

thout looking, what is the probability that it will be a red marble?
Mathematics
2 answers:
s344n2d4d5 [400]3 years ago
7 0

Answer:

Step-by-step explanation:

P(picking a red marble ) =<u> No. of  red marble</u>

                                           Total of marbles

   = 4/20 = 1/5

Nesterboy [21]3 years ago
5 0
It’s a 1/5 chance to be red
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2,11,20 find the t15
astraxan [27]

Answer:

128

Step-by-step explanation:

There is a common difference between consecutive terms, that is

11 - 2 = 20 - 11 = 9

This indicates the sequence is arithmetic with n th term

t_{n} = t₁ + (n - 1)d

where t₁ is the first term and d the common difference

Here t₁ = 2 and d = 9 , thus

t₁₅ = 2 + (14 × 9) = 2 + 126 = 128

5 0
3 years ago
338 is the same as 122 reduced by w written as an equation
Ivanshal [37]

Step-by-step explanation:

The given statement can be expressed as:

122 - w = 338

6 0
3 years ago
((2x^9)(y^n))((4x^2)(y^10))=(8x^11)(y^20)
tino4ka555 [31]

We are given

((2x^9)(y^n))((4x^2)(y^{10}))=(8x^{11})(y^{20})

Firstly, we simplify left side

Left side is

((2x^9)(y^n))((4x^2)(y^{10}))

we will make all x terms together

and y terms together

(2x^9)(4x^2)(y^n)(y^{10})

2\times 4(x^9)(x^2)(y^n)(y^{10})

8(x^9)(x^2)(y^n)(y^{10})

we can multiply left side by using exponent rule

a^m\times a^n=a^{m+n}

8(x^{9+2})(y^{n+10})

8(x^{11})(y^{n+10})

now, we can set them equal

(8x^{11})(y^{n+10})=(8x^{11})(y^{20})

Since, both sides have x,y and 8

and both are equal

so, their exponent must be equal

so, exponent of y must also be equal

we get

n+10=20

n=10................Answer



3 0
3 years ago
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Answer:

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Step-by-step explanation:

Please see attached picture for full solution.

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