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kupik [55]
3 years ago
10

7/3x = 2/3 is about solve equations with rational numbers

Mathematics
1 answer:
Leya [2.2K]3 years ago
7 0

Answer:

x=2/7

In standard form: 7x=2

implicit form:

7x-2=0

overall x=2/7 is the answer

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That would be 2.1 because if it's in decimal it's 2.1
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license plates in a particular state display two letters followed by three numbers, such as AT-887 or BB-013. How many different
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10^3*26^2
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7 0
4 years ago
For any set of n measurements, the fraction included in the interval y − ks to y + ks is at least 1 − 1 k2 . This result is know
Karo-lina-s [1.5K]

Answer:

This interval is between 1.5 and 15.5.

Step-by-step explanation:

Tchebysheff's Theorem

The Tchebysheff's Theorem can also be applied to non-normal distribution. It states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

An in general terms, the percentage of measures within k standard deviations of the mean is given by 100(1 - \frac{1}{k^{2}}).

In this question, we have that:

Mean = 8.5

Standard deviation = 3.5

The manager wants to publish an interval in which at least 75% of these values lie.

By the Tchebysheff's Theorem, at least 75% of the measures are within 2 standard deviations of the mean.

8.5 - 2*3.5 = 1.5

8.5 + 2*3.5 = 15.5

This interval is between 1.5 and 15.5.

7 0
3 years ago
Please help i don't know the answer to number 11
vivado [14]
The length of midpoint connected size is half of the size of the triangle that it's parallel to. that's means lengh of sizes are half of the original.

area of triangle with known side a, b, c
s = (a+b+c)/2
area = sqrt(s(s-a)(s-b)(s-c))

area of midpoint triangle wih side a/2, b/2, c/2
s' = (a/2+b/2+c/2)/2 = (a+b+c)/4 = s/4
area' = sqrt(s'(s'-a/2)(s'-b/2)(s'-c/2))
= sqrt(s/2(s/2-a/2)(s/2-b/2)(s/2-c/2))
= (1/4)sqrt(s(s-a)(s-b)(s-c)) = (1/4)area

answer is 1/4
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3 years ago
Where will you find the solution to a system of equations on a graph?
White raven [17]
X and Y If this does not help use khan academy
8 0
3 years ago
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