Is 5 degrees above zero a negative quantity always never or sometimes

Mathematics

1 answer:

7nadin3 [17]3 years ago

5 0

Answer:

never

Step-by-step explanation:

5 degrees above zero is never a negative, it's positive

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What is the distance from 10 To 0

kicyunya [14]

Answer:

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4 years ago

What number is 32% of 1,000

Jlenok [28]

It is 320 because if you divide 1000 by 10, you would get 100 (which would be equal to 10% of 1000) So if you multiply 100x3=300. then, 2% of 100 is 20, so you add that to 300 then you 320

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4 years ago

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas.

erastova [34]

Answer:

n=50

Step-by-step explanation:

1) 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".

$\bar X=16.10$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size

2) Solution to the problem

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

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

We need to find the degrees of freedom given by:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=\pm 1.96$

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)