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Quadrilateral ABCD is inscribed in a circle with angle measures m∠A = (11x − 8)°, m∠B = (3x2 + 1)°, m∠C = (15x + 32)°, and m∠D =

dimulka [17.4K]

Answer:

No for all of the 4 values

Step-by-step explanation:

The sum of the internal angles of a quadrilateral is always 360°, so we have that:

m∠A + m∠B + m∠C + m∠D = 360

11x - 8 + 3x2 + 1 + 15x + 32 + 2x2 - 1 = 360

5x2 + 26x - 320 = 0

Solving the quadratic function using Bhaskara's formula, we have:

Delta = 26^2 + 4*5*320 = 7076

sqrt(Delta) = 84.12

x1 = (-26 + 84.12)/10 = 5.81

x2 = (-26 - 84.12)/10 = -11.01 (this value will generate negative angles, so it is not valid).

Now, finding the angles, we have:

m∠A = 11*5.81 - 8 = 55.91°

m∠B = 3*(5.81)^2 + 1 = 102.27°

m∠C = 15*5.81 + 32 = 119.15°

m∠D = 2*(5.81)^2 - 1 = 66.51°

As all these values are different from the values shown, the answer is No for all of them.

8 0

3 years ago

The Office of Student Services at a large western state university maintains information on the study habits of its full-time st

Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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 $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .