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2 years ago

Helpppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

Mathematics

1 answer:

Citrus2011 [14]2 years ago

8 0

Answer:

Step-by-step explanation:

I D K this is a puzzle

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Simplify: Log3 27 = x can anybody help me with this?? Thank you!!

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Answer:3

Step-by-step explanation:

3 ^x= 27

X=3

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Solve this quadratic equation using the quadratic formula. 5 - 10x - 3x2 = 0

Olegator [25]

-3x^2 - 10x + 5 = 0

Quadratic formula:

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3 years ago

The graph of a polynomial is shown below. Between which two x-values does this polynomial have an extreme value?

stiv31 [10]

Answer:

C.-3 to -2

Step-by-step explanation:

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3 years ago

You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a

Karo-lina-s [1.5K]

Answer:

$573-1.690\frac{22}{\sqrt{36}}=566.8033$

$573+1.690\frac{22}{\sqrt{36}}=579.1967$

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

$\bar X=573$ represent the sample mean

$\mu$ population mean (variable of interest)

s=22 represent the sample standard deviation

n=36 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df=n-1=36-1=35$

The Confidence level is 0.90 or 90%, the significance would be $\alpha=0.1$ and $\alpha/2 =0.05$ , and the critical value for this case would be $t_{\alpha/2}=1.690$

And replacing we got:

$573-1.690\frac{22}{\sqrt{36}}=566.8033$

$573+1.690\frac{22}{\sqrt{36}}=579.1967$

And the 90% confidence interval would be between [566.8033 and 579.1967]

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4 years ago