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3 years ago

What is the greatest common factor of 60x4y7, 45x5y5, and 75x3y? 5xy 15x3y 45x3y5 75x5y7

Mathematics

2 answers:

Rasek [7]3 years ago

8 0

Answer: 15x3y

Step-by-step explanation:

We rewrite the expressions by using factorisation method of their primes. This will give:

60x4y7= 2×2×3×5×x×x3×y×y6

45x5y5= 3×3×5×x×x4×y×y4

75x3y= 3×5×5×x×x2×y

Common factors= 3 × 5 × x3 × y

= 15x3y

Pepsi [2]3 years ago

6 0

I believe it’s 5xy since all the numbers are factors of five and since each number contains the variables xy.

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Jack is a college athlete who has been weighing himself weekly on the same scale in the athletic center for the past few years.

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Answer:

a) Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=\pm 1.64$

b) $\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

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

And the margin of error is:

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And then the confidence interval is be given by:

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X= 200$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=3$ represent the population standard deviation

n=9 represent the sample size

Part a

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=\pm 1.64$

Part b

The distribution for the sample mean is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{9}}=1$

Part c

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

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

And the margin of error is:

$ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.64$

And then the confidence interval is be given by:

$200-1.64 = 198.36$

$200+1.64 = 201.64$

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