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Debora [2.8K]
3 years ago
10

Which is the lowest price ? per oz?

Mathematics
2 answers:
notka56 [123]3 years ago
7 0
Umm i not that sure but i think its a
Studentka2010 [4]3 years ago
3 0
Your answer would be a
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Help plz it is only one question
Akimi4 [234]
It would be 43.75% because 

7÷16=0.4375
0.4375×100=43.75%
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16 homework papers to grade and 14 exit tickets what's the ratio
schepotkina [342]
The ratio would be 16:14 and then simplify it to 8:7
3 0
3 years ago
Eloise has 3/4 -pound of bird seeds. Each day, she feeds the birds in her yard 1/8 -pound of seeds. For how many days will she b
bulgar [2K]

Answer:

6 days

Step-by-step explanation:

First, we make common denominators:

3 *2  =  6

_  ->     _

4 *2  =  8

and we get 6/8

6/8 divided by 1/8 will be 6 days

3 0
2 years ago
HELPPPPPP
SOVA2 [1]

Answer:

10^-99

10^-17

10^-5

10⁵

10¹⁴

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8 0
2 years ago
An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
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