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Natalka [10]
3 years ago
12

g.o. has 3 orange picks for every two green . if there are 25 picks in all , how many picks are orange

Mathematics
1 answer:
Nuetrik [128]3 years ago
8 0
3 are orange out of every 5 picks
The fraction 3/5 represents the situation
There are a total of 25 picks.
Multiply the fraction by the number of picks.

(3/5) x 25

15 of the picks were orange
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Explanation:
30 x 24 = 720
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(15 x 20)/2 = 150
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A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the followin
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Answer:

Σxi =56.61

Σxi²=196.23

Step-by-step explanation:

X be the random variable represents area of scleral lamina (mm2) from human optic nerve heads

Σxi can be obtain by adding all x values

Σxi=2.79+2.53+2.78+3.79+2.33+2.72+3.88+4.15+3.79+4.43+3.47+4.47+2.5+3.61+2.77+3.53+3.07=56.61

Σxi =56.61

Σxi² can be obtained by squaring x values and then adding them.

Σxi²=2.79²+2.53²+2.78²+3.79²+2.33²+2.72²+3.88²+4.15²+3.79²+4.43²+3.47²+4.47²+2.5²+3.61²+2.77²+3.53²+3.07²=196.233

Σxi²=196.23

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2 years ago
Which expression is equal to (7 × 2) × 2?
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Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the s
Arlecino [84]

Answer:

P(X > 0.3) = 0.3897

Step-by-step explanation:

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

Applying the central limit theorem:

\mu = p = 0.29, s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.29*0.71}{160}} = 0.0359

P( > 0.3)

This is 1 subtracted by the pvalue of Z when X = 0.3. So

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

By the Central Limit Theorem

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

Z = \frac{0.3 - 0.29}{0.0359}

Z = 0.28

Z = 0.28 has a pvalue of 0.6103

1 - 0.6103 = 0.3897

So

P(X > 0.3) = 0.3897

7 0
3 years ago
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