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3 years ago

In which case would it be most appropriate to use feet as a unit of measurement?

2 answers:

8 0

In which case would it be most appropriate to use feet as a unit of measurement? C. height of a door

The length of length of a paper clip and the width of an eyelash would be too small. It could not be measured by feet. The distance from Chicago to San Francisco would be too long and should be measured by miles.

I hope this helps! :)
~ erudite

il63 [147K]3 years ago

3 0

A height of a door because the distance from Chicago to San Francisco would be expressed as miles and the width of an eyelashes would be millimeters and the lenght of a paperclip would be centimerters.

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What's 0.7 multiplied by 9 Estimate Tell me estimation and answer in comments

kirill [66]

Estimate:6
Answer:6.3

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3 years ago

Read 2 more answers

What is the quotient? URGENT!!

jeyben [28]

Answer:

The answer is A.

Step-by-step explanation:

You have to multiply by converting the second fraction into upside down :

$\frac{4x + 1}{6x} \div \frac{x}{3x - 1}$

$= \frac{4x + 1}{6x} \times \frac{3x - 1}{x}$

$= \frac{(4x + 1)(3x - 1)}{x(6x)}$

$= \frac{12 {x}^{2} - 4x + 3x - 1}{6 {x}^{2} }$

$= \frac{12 {x}^{2} - x - 1 }{6 {x}^{2} }$

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3 years ago

I need help writing the expressions

True [87]

Answer:

Company a= 15x+10y+25z+75

Company b=20x+10y+30z+60

Step-by-step explanation:

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2 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: 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)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$