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dmitriy555 [2]
2 years ago
9

Please help me I will give you brain list

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
8 0

1 triangle=3 sticks.

2 triangle= 6 sticks.

3 triangle= 9 sticks.

4 triangles=12 sticks.

5triangles=15sticks.

10 triangles= 30 sticks.

30 triangles=90 sticks

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Quadrilateral ABCD is a square with diagonals that intersect at point E. Find angle ADB. A. 11.25º B. 22.5º C. 45º D. 90º
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IT WOULD BE "D" I BELIEVE

HOPE IM RIGHT AND THIS HELPS


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If you are driving on the freeway at 55 mph and you travel for 5 hours, how far will you have gone?
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The area of a square tile is 49 square inches. Write the length of one side of the square time as a square root and standard fro
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7 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.

6 0
2 years ago
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