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Free_Kalibri [48]
3 years ago
8

A water lily is floating in a murky pond rooted on the bottom of the pond by a stem of unknown length. The lily can be lifted 2

feet over the water and moved 6 feet to the side. What is the depth of the pond?
Mathematics
1 answer:
stiks02 [169]3 years ago
5 0
12 Feet is the depth of the pond.  I am just guessing but i think that is it.
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Please answer this correctly I have to finish the sums today as soon as possible
padilas [110]

Answer:

All of them

Step-by-step explanation:

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3 0
2 years ago
Ricardo and Tammy practice putting golf balls. Ricardo makes 47% of his putts and Tammy makes 51% of her putts. Suppose that Ric
yaroslaw [1]

Answer:

0.3821 = 38.21% probability that Ricardo makes a higher proportion of putts than Tammy.

Step-by-step explanation:

To solve this question, we need to understand the normal distribution, the central limit theorem, and subtraction of normal variables.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction of normal variables:

When we subtract normal variables, the mean of the subtraction will be the subtraction of the means, while the standard deviation will be the square root of the sum of the variances.

Ricardo makes 47% of his putts, and attempts 25 putts.

By the Central Limit Theorem, we have that:

\mu_R = 0.47, s_R = \sqrt{\frac{0.47*0.53}{25}} = 0.0998

Tammy makes 51% of her putts, and attempts 30 putts.

By the Central Limit Theorem, we have that:

\mu_T = 0.51, s_T = \sqrt{\frac{0.51*0.49}{30}} = 0.0913

What is the probability that Ricardo makes a higher proportion of putts than Tammy?

This is the probability that the subtraction of R by T is larger than 0. The mean and standard deviation of this distribution are, respectively:

\mu = \mu_R - \mu_T = 0.47 - 0.51 = -0.04

s = \sqrt{s_R^2 + s_T^2} = \sqrt{0.0998^2 + 0.0913^2} = 0.1353

This probability is 1 subtracted by the pvalue of Z when X = 0. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0 - (-0.04)}{0.1353}

Z = 0.3

Z = 0.3 has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that Ricardo makes a higher proportion of putts than Tammy.

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Step-by-step explanation:

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The answer will be 164
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(4, -5)

Step-by-step explanation:

Use the (y, -x) rule when rotating points 90 degrees clockwise on a graph.

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