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Nostrana [21]
4 years ago
6

On monday maggie ran n miles, on tuesday she ran 90% of the number of miles that she ran on monday. on wednesday she ran 50% of

the number of miles that she ran on tuesday.
move a symbol to the box and expressions to the blanks to create to equivalent expressions that represent the total number of miles maggie ran on monday, tuesday, and wednesday.

first expression: n+0.9n ____
second expression: _____n


+ x 0.45 0.45n 0.5 0.5n 1.35 1.5 2.35
Mathematics
1 answer:
Andrew [12]4 years ago
3 0

Step-by-step explanation:

On Monday maggie ran n miles.

On Tuesday, she she ran 90% of the number of miles that she ran on Monday i.e \dfrac{90}{100}\times n=0.9n.

On Wednesday, she ran 50% of the number of miles that she ran on Tuesday i.e. \dfrac{50}{100}\times 0.9n=0.45n

First expression : n+0.9n = 1.9 n

Second expression : n + 0.9 n + 0.45 n = 2.35 n

Hence, this is the required solution.

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Answer:

P(\bar X>76)

And we can use the z score formula given by:

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And if we find the z score for 76 we got:

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And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

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

Let X the random variable who represent the score of golfers. We select a sampel size of 81.

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 we want to find this probability:

P(\bar X>76)

And we can use the z score formula given by:

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

And if we find the z score for 76 we got:

z =\frac{76-75}{\frac{4.5}{\sqrt{81}}}= 2

And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

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