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My name is Ann [436]
2 years ago
11

Solve for x using Logarithm4^x+6^x =9^x​

Mathematics
1 answer:
vitfil [10]2 years ago
7 0
4^x + 6^x = 9^x

4^x + 6^x - 9^x = 0

2^2x + (2×3)^x - 3^2x = 0

2^2x + 2^x×3^x - 3^2x = 0

(⅔)^2x + (⅔)^x - 1 = 0

let t = ( ⅔)^x

t² + t - 1 = 0

t = (-1+√5 ) / 2

(⅔)^x = (-1+√5 ) / 2

x = Log base ⅔ ((-1+√5 ) / 2 )


x = 1.18681
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2 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
hoa [83]

Answer:

Obese people

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

5 0
2 years ago
When the expression $-2x^2-20x-53$ is written in the form $a(x+d)^2+e$, where $a$, $d$, and $e$ are constants, then what is the
Sladkaya [172]
We start with 2 x^{2} -20x-53 and wish to write it as a(x+d) ^{2} +e

First, pull 2 out from the first two terms: 2( x^{2} -10x)-53

Let’s look at what is in parenthesis. In the final form this needs to be a perfect square. Right now we have x^{2} -10x and we can obtain -10x by adding -5x and -5x. That is, we can build the following perfect square: x^{2} -10x+25=(x-5) ^{2}

The “problem” with what we just did is that we added to what was given. Let’s put the expression together. We have 2( x-5) ^{2}-53 and when we multiply that out it does not give us what we started with. It gives us 2 x^{2} -20x+50-53=2 x^{2} -20x-3

So you see our expression is not right. It should have a -53 but instead has a -3. So to correct it we need to subtract another 50.

We do this as follows: 2(x-5) ^{2}-53-50 which gives us the final expression we seek:

2(x-5) ^{2}-103

If you multiply this out you will get the exact expression we were given. This means that:
a = 2
d = -5
e =  -103

We are asked for the sum of a, d and e which is 2 + (-5) + (-103) = -106


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3 years ago
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matrenka [14]

Answer:

18 combinations

Step-by-step explanation:

thank u

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2 years ago
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