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Andreyy89
2 years ago
11

Root of 3 time the root of 5

Mathematics
1 answer:
zimovet [89]2 years ago
3 0

Answer:

root of 15 or in decimal form 3.87 (rounded)

Step-by-step explanation:

You just multiply both together

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The answer is (c) 368.96

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Write <br> 6<br> :<br> 9<br> :: <br> 2<br> :<br> 3<br> as a fraction proportion.
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6/9 (2/3) and 2/3. Hope this helps you.
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I need help please...............
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Step-by-step explanation:

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3 years ago
In college​ basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points ins
denis-greek [22]

Answer:

a) By the Central Limit Theorem, the distribution would be approximately normal, with mean \mu = 0.45 and standard deviation s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497

b) 4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered surprising, so yes, it would be surprising for him to make only 25 of these shots.

Step-by-step explanation:

To solve this question, we are going to need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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 proportions, we have that the mean is \mu = p and the standard deviation is s = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.45, n = 100

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean \mu = 0.45 and standard deviation s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497

b)This is Z when X = 0.25.

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

By the Central Limit Theorem

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

Z = \frac{0.25 – 0.45}{0.0497}

Z = -4.02

4.02 standard deviations below the mean.

c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered

surprising, so yes, it would be surprising for him to make only 25 of these shots.

3 0
3 years ago
Determine the scale factor for each dilation. Determine whether the dilation is an enlargement, reduction, or congruence transfo
scoray [572]

General Idea:

W hen we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)

Whenk>1, then P' is enlargement of P

When 0 < k < 1, then P' is reduction of P

Applying the concept:

In the diagram given A'B'C' is enlargement of ABC. So the correct option will be option A. k = 2, enlargement

6 0
2 years ago
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