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seropon [69]
3 years ago
13

Find the perimeter of the polygon. Please help

Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

Answer:

the answer is 96 it's 12 x 8

creativ13 [48]3 years ago
5 0

Answer:

EASY 96

Step-by-step explanation:

just add all sides together 12+12+12+12+12+12+12+12=96

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alexandr402 [8]
120 hope this helps ;)
4 0
3 years ago
Find the square root of 1.69? show work ...?
andreyandreev [35.5K]
Maybe something like this:
1.69 = 169 / 100
√ 1.69 = √ ( 169 /100 ) = √169 / √ 100 = 13 / 10 = 1.3
Thank you.
3 0
3 years ago
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What is $0.75 of $16.40​
Igoryamba

Answer:

4.5732 %

Step-by-step explanation:

6 0
2 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
Please help
oksian1 [2.3K]

Answer:

1: 11√3 - 7√6

2: 11√3 - 7√6

3: -9

4:12

Step-by-step explanation:

To add radicals they need to have the same radical part for the first one we have

7√3- 4√6 + √48 - √54

We can simplify the last two into 4√3 and 3√6

So we have 7√3 - 4√6 + 4√3 - 3√6

adding similar radicals we get

11√3 - 7√6

For the second one we have 11√3 - 7√6

There's nothing we can do from here so keep that as your answer

This one is quite easy -3√9

square root of 9 is 3

so we have -3*3 which is -9

next is

4√9

same deal as the one before

3*4=12

5 0
2 years ago
Read 2 more answers
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