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Doss [256]
3 years ago
12

Find the value of -8 + 9.2 +-3.​

Mathematics
2 answers:
Airida [17]3 years ago
8 0

Answer:

-1.8

Step-by-step explanation:

Because,

-8 + 9.2 +-3

=> 9.2 - 8 - 3 (Commutative property)

=> 9.2 - 11 (Because subtraction is done first, and a negative + a negative = a negative)

=> -1.8 will be the final answer

Hope this helped, Please mark me as brainliest, Thank you!

skelet666 [1.2K]3 years ago
6 0

Answer:

-1.8

Step-by-step explanation:

move the numbers around a bit because there is no multiplication so 9.2-8+-3

then take out the plus so it doesn't confuse anyone 9.2-8-3

then add 8+3 which is 11 but remember to put the negative back

9.2 - 11

which gives you -1.8

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Suppose 46% of politicians are lawyers. If a random sample of size 662 is selected, what is the probability that the proportion
Svet_ta [14]

Answer:

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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}}

Suppose 46% of politicians are lawyers.

This means that p = 0.46

Sample of size 662

This means that n = 662

Mean and standard deviation:

\mu = p = 0.46

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*0.54}{662}} = 0.0194

What is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%?

p-value of Z when X = 0.46 + 0.04 = 0.5 subtracted by the p-value of Z when X = 0.46 - 0.04 = 0.42. So

X = 0.5

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.46}{0.0194}

Z = 2.06

Z = 2.06 has a p-value of 0.9803

X = 0.42

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

Z = \frac{0.42 - 0.46}{0.0194}

Z = -2.06

Z = -2.06 has a p-value of 0.0197

0.9803 - 0.0197 = 0.9606

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

8 0
3 years ago
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