(x + 5)2 + (y - 7)2 = 36, its radius is

1 answer:

5 0

6 because the 36 is r² where r is the radius, r² = 36; we have to isolate the r so it's by its self, so square root both sides, r = 6

You might be interested in

The radius of the cone is 7 in and y = 25 in. What is the volume of the cone in terms of π? A cone with a right triangle formed

Lady bird [3.3K]

Answer:

142.1π in³

Step-by-step explanation:

Given that:

The radius (r) = 7 in

The slant height (y) = 25 in

Then the height (x) can be determined by using the Pythagoras rule:

y² = x² + r²

25² = x² + 7²

125 = x² + 49

125 - 49 = x²

x² = 76

x = √76

x = 8.7

The formula for the volume of a cone is;

= 1/3 πr²h

where;

height(h) is calculated as "x" from above = 8.7

Then;

= 1/3 × π × (7 in)² × 8.7 in

= 142.1π in³

5 0

3 years ago

Read 2 more answers

Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The p

Ymorist [56]

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .