Marvin says that all rhombuses are squares. Aretha says that all squares are rhombuses. who is correct?
1 answer:
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Let's say our first integer is "a".
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
![\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20times%20the%20first%7D%7D%7B3a%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B26%20less%20than%20twice%20the%20sum%20of%20the%20others%7D%7D%7B2%5B~%28a%2B2%29%2B%28a%2B4%29~%5D~~~-26%7D%0A%5C%5C%5C%5C%5C%5C%0A3a%3D2%5B~2a%2B6~%5D-26%5Cimplies%203a%3D4a%2B12-26%5Cimplies%203a%3D4a-14%0A%5C%5C%5C%5C%5C%5C%0A0%3Da-14%5Cimplies%2014%3Da)
what are the other two consecutive integers? well, a + 2 and a + 4.
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
what are the other two consecutive integers? well, a + 2 and a + 4.
Answer:
0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.
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 and standard deviation
, the z-score of a measure X is given by:
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 and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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 and standard deviation
Proportion of 0.6
This means that
Sample of 46
This means that
Mean and standard deviation:
Probability of obtaining a sample proportion less than 0.5.
p-value of Z when X = 0.5. So
By the Central Limit Theorem
has a p-value of 0.0838
0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.