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lorasvet [3.4K]
2 years ago
9

. Зn – 1 = 8 Pls tell me what n equals lol

Mathematics
2 answers:
Naddika [18.5K]2 years ago
8 0

Answer:

Зn – 1 = 8

3n=9

n=3

Hope This Helps!!!

Nana76 [90]2 years ago
4 0

Answer:

n=3

Step-by-step explanation:

add 1 to both sides

then divide both sides by 3

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What is the best approximation for the input value when f(x)=g(x)?
Lostsunrise [7]

Answer:

x=0 and x=1.

Step-by-step explanation:

If we have to different functions like the ones attached, one is a parabolic function and the other is a radical function. To know where f(x)=g(x), we just have to equalize them and find the solution for that equation:

x^{2}=\sqrt{x} \\(x^{2} )^{2}=(\sqrt{x} )^{2}\\x^{4}=x\\x^{4}-x=0\\x(x^{3}-1)=0\\

So, applying the zero product property, we have:

x=0\\x^{3}-1=0\\x^{3}=1\\x=\sqrt[3]{1}=1

Therefore, these two solutions mean that there are two points where both functions are equal, that is, when x=0 and x=1.

So, the input values are  x=0 and x=1.

8 0
3 years ago
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Does anyone know this one?
iogann1982 [59]
It's A.

18 correlates with $350 and 12 correlates with $520.
That crosses off B and D.

18 + 12 days combined equals 30.
That excludes C and D.

A is your answer! :D
6 0
3 years ago
What does m equal in this equation? <br> 5m+2(m+1)=23
barxatty [35]
5m+2m+2 = 23
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8 0
2 years ago
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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
2 years ago
URGENT HELP PLEASE.
34kurt

Answer:

i hope this works. Feel better soon!

Step-by-step explanation:

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4 0
2 years ago
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