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babymother [125]
3 years ago
9

Read the following statement: All prime numbers are odd. Which of the following choices is a counterexample to the statement?

Mathematics
1 answer:
lana66690 [7]3 years ago
5 0
All prime numbers are odd.
A counter-example is one that will prove this statement to be false.

ur answer is : 2....it is 2 because 2 is a prime number and it is even...in fact, it is the only even number that is a prime number.


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If you choose each randomly, what is the probability that you choose a peach pie and coffee? Write as a simplified fraction.
emmasim [6.3K]

Answer:

1\4

Step-by-step explanation:

i did it out of 100 i choose 25/100

then id you simplified that 25 goes into 100 4 times 25 goes into int self one time so the answer is 1/4

4 0
2 years ago
In this rectangular box, E F = 16, FD = 5, and DB = 30. Find A F.
DENIUS [597]
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Hope this helped :D
7 0
3 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
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Lapatulllka [165]

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(1/2)x3-(1/3)x2

3/6-2/6

1/6

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3 years ago
Find the area of a regular hexagon with a 48- inch perimeter
solmaris [256]

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7 0
3 years ago
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