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Savatey [412]
3 years ago
11

Danika signs up to work for 3 1/2 hours at the science fair. If each work shift is 3/4 hour, how many shifts will Danika work?

Mathematics
2 answers:
forsale [732]3 years ago
5 0

namely, how many times does 3/4 go into 3½?  Let's firstly convert the mixed fraction to improper fraction.

\bf \stackrel{mixed}{3\frac{1}{2}}\implies \cfrac{3\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{7}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{2}\div \cfrac{3}{4}\implies \cfrac{7}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{\stackrel{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{3}\implies \cfrac{14}{3}\implies 4\frac{2}{3}

Liula [17]3 years ago
4 0

Answer:

Danika signed up to work for 3.5 hours at the science fair, if each work shift is 3/4 hour = 0.75 hours, the amount of shifts will be:

Shifts = 3.5/0.75 = 4,66 ≈ 5 shifts. Given that shifts have to be an integer number, we need to round the result to the nearest integrer.

Therefore, Danika will work 5 shifts.

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Answer:

Let X the random variable who represents the file sizeof music. We know the following info:

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Step-by-step explanation:

Previous concepts

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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".

Let X the random variable who represents the file sizeof music. We know the following info:

\mu =2.3,\sigma =3.25

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this 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 we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

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Answer:

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Step-by-step explanation:

hope this helps:)

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