In the xy plane, a quadrilateral has vertices at (-1, 4), (7,4), (7,5), and (-1. 5). What is the perimeter of the quadrilateral?
(A) 17 (B) 18 (C) 19 (1) 32 (E) 34
1 answer:
Answer:
(B) 18.
Step-by-step explanation:
We are asked to find the perimeter of a quadrilateral with vertices at (-1, 4), (7,4), (7,5), and (-1. 5).
First of all, we will draw vertices of quadrilateral on coordinate plane and connect the vertices as shown in the attached photo.
We can see that our quadrilateral is a parallelogram, whose parallel sides are equal.
Therefore, the perimeter of the given quadrilateral is 18 units.
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Answer:
50.34% probability that the sample mean would differ from the true mean by less than 1.1 months
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 normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 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 and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months
This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So
X = 120.1
By the Central Limit Theorem
has a pvalue of 0.7517
X = 117.9
has a pvalue of 0.2483
0.7517 - 0.2483 = 0.5034
50.34% probability that the sample mean would differ from the true mean by less than 1.1 months