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marin [14]
3 years ago
8

It took Kelly 5 2/3 hours to fly from City A to City B then it took her another 2 3/5hours to fly from City B toCity C. Kelly fl

ew a total how many hours?
Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0

Answer:

eto na po

Step-by-step explanation:

5 2/3 × 2 3/5 = 5 6/15 answer

bogdanovich [222]3 years ago
5 0
Just add 5 2/3 to 2 3/5!
the answer is 8 4/15
workings: you can convert the fractions to an improper fraction and add them, it would be easier!
5 2/3 > 17/3 (eq 1)
2 3/5 > 13/5 (eq 2)
after conversion, you must make the denominator the same, the common multiple is 15
take eq 1 multiply by 5 > 85/15
take eq 2 multiply by 3 > 39/15
85/15 + 39/15 = 124/15!!
124/15 convert back to mixed fraction > 8 4/15
hope it helped :D
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3 years ago
Can anyone pls help with this ASAP!!! :((
MrRissso [65]

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The length of a phone conversation is normally distributed with a mean of 4 minutes and a standard deviation of 0.6 minutes. Wha
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6 0
3 years ago
An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was x = 119.6 ounc
Evgen [1.6K]

Answer:

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}})

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

c. 1.30 ounces.

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

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}})

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

c. 1.30 ounces.

6 0
3 years ago
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