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Vera_Pavlovna [14]
3 years ago
6

Pierre is running 26 and 2/10 miles in the marathon. he has ran 3/4 of the way. How far has he run?

Mathematics
1 answer:
vredina [299]3 years ago
3 0
Distance=(3/4) * (26  2/10  miles)

26  2/10  miles=26  1/5 miles=(5*26+1) / 5 miles=131/5 miles.

Therefore:

distance=(3/4) * (26  2/10 miles)=(3/4) * (131 /5 miles)=
=(131 *3) / (4*5) miles=393/20 miles= (19*20+13)20 miles=19  13/20 miles.


he has run 393/20 miles    or    19  13/20  miles

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look up desmos graphing calculator and it graphs all of your equations for you. I used it all through high school.

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Please answer with clear instructions so that i can apply this to other questions.
kvv77 [185]

Answer:

10 terms

Step-by-step explanation:

equate the sum formula to 55 and solve for n

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n(n + 1) = 110 ← distribute parenthesis on left side

n² + n = 110 ( subtract 110 from both sides )

n² + n - 110 = 0 ← in standard form

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the factors are + 11 and - 10 , since

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(n + 11)(n - 10) = 0 ← in factored form

equate each factor to zero and solve for n

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number of terms which sum to 55 is 10

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2 years ago
What is the sum?
OleMash [197]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small files. Suppose the true me
m_a_m_a [10]

Answer:

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:

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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.

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:

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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:

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