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3 years ago

What is the center of the equation X^2+y^2-4x-8y-16=0

1 answer:

6 0

x² + y² - 4x - 8y - 16 = 0
+ 16 + 16
x² + y² - 4x - 8y = 16
x² - 4x + y² - 8y = 16
(x² - 4x + 4) + (y² - 8y + 16) = 16 + 4 + 16
(x² - 2x - 2x + 4) + (y² - 4y - 4y + 16) = 20 + 16
[x(x) - x(2) - 2(x) - 2(-2)] + [y(y) - y(4) - 4(y) - 4(4)] = 36
[x(x - 2) - 2(x - 2)] + [y(y - 4) - 4(y - 4)] = 36
(x - 2)(x - 2) + (y - 4)(y - 4) = 36
(x - 2)² + (y - 4)² = 36
36 36 36
¹/₃₆(x - 2)² + ¹/₃₆(y - 4)² = 1
Center: (2, 4)

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Which of the following points lies on the circle whose center is at the origin and whose radius is 10?

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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

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Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

$Z = \frac{X - \mu}{s}$

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

$Z = \frac{X - \mu}{s}$

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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3 years ago

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Step-by-step explanation:

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3 years ago

A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How far does the ju

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Answer:

x = 9.5 m

Step-by-step explanation:

Given:

This problem is based on 2D motion kinematics equations:

$V_0 = 12 m/s$

$\theta = 20^0$

$g=9.8 m/s^2$

At maximum height:

$V_y = 0$

Components of initial velocity:

$V_0x = V_0 cos (\theta) =(12 )(cos (20^0)) = 11.3 m/s$

$V_0x = V_0 sin (\theta) = (12)(sin (20^0)) = 4.10 m/s$

To find time to reach maximum height:

$V_y = V_0y -gt$

Plugging in the known values:

0 = 4.10 - 9.8t

$t = \frac{-4.10}{-9.8}$

$t = 0.418 s$

Total time = 2t = 2(0.418) = 0.836 s

to find horizontal distance covered:

$x = (V_0x)(t)\\x = (11.3)(0.836)\\x = 9.5 m$

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3 years ago