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Katen [24]
3 years ago
13

The denominator of a fraction is one more than the numerator. If 3 is subtracted to

Mathematics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

\frac{4}{5}

Step-by-step explanation:

let the original fraction be

\frac{n}{n+1} ( with denominator 1 more than numerator )

Subtracting 3 from numerator/ denominator gives

\frac{n-3}{n+1-3} = \frac{n-3}{n-2}

Given that the fraction is now equivalent to \frac{1}{2} , then

\frac{n-3}{n-2} = \frac{1}{2} ( cross- multiply )

2(n - 3) = n - 2, distribute left side

2n - 6 = n - 2 ( subtract n from both sides )

n - 6 = - 2 ( add 6 to both sides )

n = 4

and n + 1 = 4 + 1 = 5

Thus original fraction is \frac{4}{5}

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3 years ago
What number must you add to complete the square x^-8x=39
notsponge [240]

Answer:

Given equation: x^2-8x =39

when we complete the square , we take half of the value of 8 , then square it, and added to the left sides, we get;

x^2-8x+4^2 = 39 +4^2

∵8 is the value (\frac{8}{2})^2

Notice that, we add this both sides so that it maintains the equality.

then;

x^2-8x+4^2 = 39 +4^2

(x-4)^2 = 39 + 16   [ (a-b)^2 = a^2-2ab+b^2 ]

Simplify:

(x-4)^2 =55

The number must be added to complete the square is, 4^2 = 16



3 0
3 years ago
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Can someone help with this?! The answer choices are 1, 1.3, 1.45, and 1.9!
Zarrin [17]
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4 years ago
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99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest. true or false
tester [92]

The statement that 99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest is false.

A confidence interval (CI) is essentially a range of estimates for an unknown parameter in frequentist statistics. The most frequent confidence level is 95%, but other levels, such 90% or 99%, are infrequently used for generating confidence intervals.

The confidence level is a measurement of the proportion of long-term associated CIs that include the parameter's true value. This is closely related to the moment-based estimate approach.

In a straightforward illustration, when the population mean is the quantity that needs to be estimated, the sample mean is a straightforward estimate. The population variance can also be calculated using the sample variance. Using the sample mean and the true mean's probability.

Hence we can generally infer that the given statement is false.

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2 years ago
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Here you go!! I hope this helps you!!

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