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pshichka [43]
3 years ago
8

- (09.02)

Mathematics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

1/3

Step-by-step explanation:

The only way it could be less than 9 is if the second cube is either 1 or 2. The probability of either of those outcomes is 2/6, simplified to be 1/3

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Aleks04 [339]

Answer:

x1=8

x2=1/5

Step-by-step explanation:

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3 years ago
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Helppppppppp me plz I rlly. We’d help
rodikova [14]

Answer:

c = 4,9

d = 4,6

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Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of resi
Ymorist [56]

Answer:

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

Step-by-step explanation:

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

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}}

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that p = 0.18, n = 100

So, by the Central Limit Theorem:

\mu = 0.18, s = \sqrt{\frac{0.18*0.82}{100}} = 0.0384

How likely is the resulting sample proportion to be between 0.066 and 0.294?

This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So

X = 0.294

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

By the Central Limit Theorem

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

Z = \frac{0.294 - 0.18}{0.0384}

Z = 2.97

Z = 2.97 has a pvalue of 0.9985

X = 0.066

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

Z = \frac{0.066 - 0.18}{0.0384}

Z = -2.97

Z = -2.97 has a pvalue of 0.0015

0.9985 - 0.0015 = 0.997

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

3 0
3 years ago
24/16 in simplist form
bonufazy [111]
1  1/2 
Hope I helped! (^_^)/
7 0
4 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Cfrac%7By%5E2%7D%7B36%7D%20-1" id="TexFormula1" title="\frac{y^2}{36} -1" alt="\frac{y^2}{36
liraira [26]

Answer:

that's my answer

Step-by-step explanation:

i hope it helps

5 0
3 years ago
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