All categories

4 years ago

What is the measure of each interior angle of a regular polygon with 5 sides?

Mathematics

2 answers:

Sedaia [141]4 years ago

5 0

108 degrees because of the equation 180(n-2)/n. 180(5-2)/5= 540/5= 108.

Ann [662]4 years ago

3 0

ANSWER

108°

EXPLANATION

The measure of each interior angle of a regular polygon is given by

$\theta = \frac{(n - 2) \times 180}{n}$

where n refers to the number of sides of the polygon.

From the question we have n=5 in this case.

We substitute to obtain;

$\theta = \frac{(5 - 2) \times 180}{5}$

$\theta = \frac{3\times 36}{1}$

This simplifies to:

$\theta = 108 \degree$

Hence each interior angles of a regular polygon with 5 - sides is 108°

You might be interested in

Tricia read 1/4 of her book on monday. on tuesday, she read 36% of the book. on wednesday, she read 0.27 of the book. she finish

8090 [49]

87% is the answer to the question

7 0

4 years ago

n a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of inches and a stand

kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

The complete question is: In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = the heights of men in the 20-29 age group

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean height = 67.5 inches

$\sigma$ = standard deviation = 2 inches

So, X ~ Normal( $\mu=67.5, \sigma^{2}=2^{2}$ )

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

P(X < 67 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{67-67.5}{2}$ ) = P(Z < -0.25) = 1 - P(Z $\leq$ 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X $\leq$ 67 inches)

P(X < 71 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{71-67.5}{2}$ ) = P(Z < 1.75) = 0.9599

P(X $\leq$ 67 inches) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{67-67.5}{2}$ ) = P(Z $\leq$ -0.25) = 1 - P(Z < 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

P(X > 71 inches) = P( $\frac{X-\mu}{\sigma}$ > $\frac{71-67.5}{2}$ ) = P(Z > 1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.9599 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

7 0

3 years ago

The mayor is interested in finding a 95% confidence interval for the mean number of pounds of trash per person per week that is

Hatshy [7]

Answer:

A. Normal

B. 30.1, 32.9

C. 95, 5

Step-by-step explanation:

A. The sampling distribution follows a normal distribution

Given that the sample size is large, we have that the sample distribution follows a normal distribution according to the central limit theorem

B. The 95% confidence interval is given as follows;

$CI=\bar{x}\pm z \times \dfrac{s}{\sqrt{n}}$

The number of residents in the study, n = 120 residents

The sample mean, $\overline x$ = 31.5 pounds

The standard deviation, s = 7.8 pounds

The z-value for 95% confidence level, z = 1.96

Therefore, we get;

C.I. = 31.5 ± 1.96 × 7.8/√(120)

The 95% C.I. ≈ 30.1 ≤ $\overline x$ ≤ 32.9

Therefore, we have that with 95% confidence, the population mean number of pounds per person per week is between 30.1 and 32.9

C. Therefore, according to the central limit theorem, about 95 percent of the groups of 120 will contain the true population mean number of pounds of trash generated per person per week and about 5 percent will not contain the true population mean number of pounds of trash generated per person per week.

8 0

3 years ago

5. A 12 sided diɛ is rolled. The set of equally

professor190 [17]

Answer:

Step-by-step explanation:

5 0

3 years ago

Please help me !!!! Will mark brainliest

stich3 [128]

Answer:

The value of x is 108°