What is the standard form of the equation of the circle to the right?

Please help the picture is

asambeis [7]

The walls are vertical so the angle opposite X would be a right angle which is 90 degrees.

X would be 180 - 90 - 35 = 55 degrees

X and 2y - 5 are complementary angles which add together to equal 90:

2y - 5 + 55 = 90

Simplify:

2y +50 = 90

Subtract 50 from both sides:

2y = 40

Divide both sides by 2:

y = 20

7 0

3 years ago

Read 2 more answers

A shower uses 12 gallons of water in 3 minutes. Complete the table and graph.

Advocard [28]

That would be 4 gallons every minute!

Also: I don't see a graph..? Is there supposed to be one attached?

Good luck!

3 0

3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your ans

Alenkinab [10]

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

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 $\mu$ 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .