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IgorLugansk [536]
2 years ago
10

Six pairs of socks cost as much as 1 coat, 2 pairs of jeans cost as much as 3 pairs of shoes, and 4 pairs of socks cost as much

as one pair of jeans. How many coats could i exchange for 64 pairs of socks.
Options: A:4 B:1 C:2 D:3 E: None of these

I do not know how to work this question out.
Please Help me
Mathematics
1 answer:
REY [17]2 years ago
7 0
(1) 6 pairs of socks = 1 coat
(2) 2 pairs of jeans = 3 pairs of shoes
(3) 4 pairs of socks = 1 pair of jeans

the last two points are irrelevant becuase we just need socks and coats. 
6 pairs of socks are equal to 1 coat. 
we need 64 pairs of socks. 
so, 64 ÷ 6 = 10.67

so we'll need 10.67 coats for 64 pairs of socks. 

i'm not sure if you've typed up the correct numbers becuase i don't think the answer would be expressed as a decimal. 

unless we say that we need 11 pairs of coats? 

perhaps reread the question again and make sure you've typed it up correctly
xD i hope i've helped. 
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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

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

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

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

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

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Answer:

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Step-by-step explanation:

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Question:

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Answer:

(a) 6 choices

(b) 3 Choices

Step-by-step explanation:

Given:

3 Colors

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(a) Order of the choices taken into consideration :

When we are picking  the first colour there are 3 color choice.

When we are picking second colour there are 2 choice and so on.

Therefore, total choices are = 3*2 = 6 choices

(b) Order of the choices not taken into consideration :

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Complete Question: Check the file attached to this document to see the complete question

Answer:

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Step-by-step explanation:

Counting properly from the diagram attached to the question,

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The total number of trials she made = 50

Probability that it takes fewer than 7 shots to get her first successful shot, P(X < 7) = (number of trials it took to make less than 7 shots)/(Total number of trials)

P(X < 7) = 38/50

P(X < 7) = 0.76

8 0
2 years ago
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