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mihalych1998 [28]
4 years ago
14

1 1/6 z = 1/5 solve for z, please show work?

Mathematics
2 answers:
geniusboy [140]4 years ago
5 0
The answer would be 6/35 

Alika [10]4 years ago
4 0
Hello!

You can solve this algebraically

1 1/6 z = 1/5

Convert the mixed number into an improper fraction

7/6 z = 1/5

Multiply both sides by 6

7z = 6/5

Divide both sides by 7

z = 6/35

The answer is 6/35

Hope this helps!
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Step-by-step explanation:

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Suppose that a risk-free investment will make three future payments of $250 in 1 year, $250 in 2 years, and $250 in 3 years. Ins
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Find the equation of the line in the xy-plane that contains the point (9, 3) and that is parallel to the line y = 6x – 4.
aliya0001 [1]
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y-yo=m(x-xo)
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Express your answer as a single term<br>​
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Step-by-step explanation:

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3 years ago
An integer N is to be selected at random from {1, 2, ... , (10)3 } in the sense that each integer has the same probability of be
Andrej [43]

Answer:

Probability of N Divisible by 3 - 0.33

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Step-by-step explanation:

Given data:

Integer N {1,2,.....10^3}

Thus total number of ways by which 1000 is divisible by 3 i.e. 1000/3 = 333.3

Probability of N divisible by 3 {N%3 = 0 } = \frac{333.3}{1000} = 0.33

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total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857

Probability of N divisible by 7 {N%7 = 0 } = \frac{142.857}{1000} = 0.413

total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667

Probability of N divisible by 15 {N%15 = 0 } = \frac{66.667}{1000} = 0.066

total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52

Probability of N divisible by 105 {N%105 = 0 } = \frac{9.52}{1000} = 0.0095

similarly for N is selected from 1,2.....(10)^k where K is  large then the N value. Therefore effect of k will remain same as previous part.

3 0
3 years ago
Read 2 more answers
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