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2 years ago

The image shows a circle centered at point O angle AOB Measures at 42 degrees.

Mathematics

1 answer:

trasher [3.6K]2 years ago

7 0

The measure of angle ACB for circle centered at point O is 21°

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

∠AOB = 2 * ∠ACB (angle at center is twice the angle at circumference)

42 = 2 * ∠ACB

∠ACB = 21°

The measure of angle ACB for circle centered at point O is 21°

Find out more on equation at: brainly.com/question/2972832

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3 years ago

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2 years ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

= $P(- 0.7< Z< 2.1)$

= $(Z< 2.1) - (Z< -0.7)$

= 0.9821 - .2420

= 0.7401

3 0

3 years ago