Quadrilateral OPQR is inscribed in circle N, as shown below. Which of the following could be used to calculate the measure of ∠O
PQ?
Circle N is shown with a quadrilateral OPQR inscribed inside it. Angle O is labeled x plus 16. Angle P is not labeled. Angle Q is labeled 6x minus 4. Angle R is labeled 2x plus 16.
Group of answer choices
m∠OPQ + (2x + 16)° = 180°
m∠OPQ = (6x − 4)° + (2x + 16)°
m∠OPQ + (x + 16)° + (6x − 4)°= 360°
m∠OPQ = (x + 16)° + (6x − 4)°
Circle N is shown with a quadrilateral OPQR inscribed inside it. Angle O is labeled x plus 16. Angle P is not labeled. Angle Q is labeled 6x minus 4. Angle R is labeled 2x plus 16.
Group of answer choices
m∠OPQ + (2x + 16)° = 180°
m∠OPQ = (6x − 4)° + (2x + 16)°
m∠OPQ + (x + 16)° + (6x − 4)°= 360°
m∠OPQ = (x + 16)° + (6x − 4)°
1 answer:
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Answer:
58.32% probability that a randomly selected application will report a GMAT score of less than 600
93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600
98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600
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 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.
In this problem, we have that:
What is the probability that a randomly selected application will report a GMAT score of less than 600?
This is the pvalue of Z when X = 600. So
has a pvalue of 0.5832
58.32% probability that a randomly selected application will report a GMAT score of less than 600
What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?
Now we have
This is the pvalue of Z when X = 600. So
has a pvalue of 0.9351
93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600
What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?
Now we have
has a pvalue of 0.9838
98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600
Answer:
First day: 45x +20y = 875
Second day: 25x + 40y = 775
Adult ticket price: $15
Step-by-step explanation:
See paper attached. (:
