Ask question

Ask question

All categories

3 years ago

11

What is the area of a regular octagon with a side length of 4.6 meters and a length from the center to a vertex of 6 meters? HEL

P ASAP BRAINLY AND 20 POINTS
A) 30.7 m^2
B) 36.8 m^2
C) 101.2 m^2
D) 202.4 m^2

1 answer:

Shtirlitz [24]3 years ago

6 0

<span>What is the area of a regular octagon with a side length of 4.6 meters and a length from the center to a vertex of 6 meters?

Good Morning

The Correct Answer is c) 101.2

Hope I Helped

-Nullgaming650

</span>

You might be interested in

Please help!!! Thank you!!!

Bess [88]

Answer:

1. 180° ∑triangle angles=180°

2. 2x

3. 52°

5 0

3 years ago

Which unit would I use to measure the circumference of a birthday cake?

Helen [10]

Depends on how rich you are
if you are very rich, use meters (or yards)
if you are kinda rich, use feet
if you are average, use centimeters (or inches)
if poor, use milimeters

6 0

3 years ago

What’s the answer to -x= -14 How do you do it?

denis23 [38]

Step-by-step explanation:

multiply both sides by -1

x = 14

5 0

3 years ago

Geometric Series assistance

Levart [38]

we have been asked to find the sum of the series

$\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}$

As we know that a geometric series has a constant ratio "r" and it is defined as

$r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}$

The first term of the series is $a_1=\left(\frac{1}{3}\right)^{1-1}=1$

Geometric series sum formula is

$S_n=a_1\frac{1-r^n}{1-r}$

Plugin the values we get

$S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}$

On simplification we get

$S_5=\frac{121}{81}$

Hence the sum of the given series is $\frac{121}{81}$

5 0

3 years ago

Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose

spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The home team therefore wins 50% of its games

This means that $p = 0.5$

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that $n = 80$ and, by the Central Limit Theorem:

$\mu = p = 0.65$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559$

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

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

By the Central Limit Theorem

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

$Z = \frac{0.65 - 0.5}{0.0559}$

$Z = 2.68$

$Z = 2.68$ has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

7 0

3 years ago