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2 years ago

Circle o has a circumference of 36π cm. what is the length of the radius, r? 6 cm 18 cm 36 cm 72 cm

2 answers:

6 0

The circumference of a circle is C = 2<span>πr
Given C = 36</span><span>π, we can solve for the radius r:
r = C/2</span>π = 36π/2<span>π = 36/2 = 18 cm (ANSWER)
</span>

iogann1982 [59]2 years ago

5 0

the answer is 18 i hope this helps

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There are 25 garden boxes in a park. Each box has 162 flowers. What is the total number of flowers in the boxes?2,750 flowers3,0

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The answer is 4050 flowers in 25 boxes

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3 years ago

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There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7. 5 and the standard d

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The margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

<h3>What is the margin of error?</h3>

The margin of error can be defined as the amount of random sampling error in the results of a survey. It is given by the formula,

$\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\frac{\sigma^{2}}{n}}$

$\text{MOE}$ = margin of error

$\gamma$ = confidence level

$z_{\gamma}$ = quantile

σ = standard deviation

n = sample size

As it is given that the sample size of the survey is 324, while the standard deviation of the survey is 1.6.

We know that the value of the z for 95% confidence interval is 1.96. Therefore, using the formula of the standard of error we can write it as,

$\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\dfrac{\sigma^{2}}{n}}\\\\\text{MOE}_{\gamma}=1.96 \times \sqrt{\dfrac{1.6^{2}}{324}}\\\\\text{MOE}_{\gamma}=0.1742$

Hence, the margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

Learn more about Margin of Error:

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2 years ago

What type of graph organizes data into 4 groups of equal size, and is often used to compare two sets of data.

aliina [53]

The plot that organizes the data into 4 groups of equal sizes is box and whisker plot.

The image below shows a box and whisker plot. Following are the elements of box and whisker plot:

Minimum = This is the smallest value of the data set

Q1 = First (Lower) Quartile of the data set. 25% of the data values lie below this point

Q2 = Second Quartile or Median. This is the central value so 50% of the data values lie below this point

Q3 = Third (Upper) Quartile of the data set. 75% of the data values lie below this point.

Maximum = This is the maximum value of the data set.

Based on box and whisker plot we can compare two or more sets of data by comparing the spread of the data. We can also directly observe from the box and whisker plot if the data is uniform, normal or skewed. Using box and whisker plot we can also visualize any outliers that may be in the data.

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3 years ago

A bean salad has 20 beans in all. It is made up of:

vlabodo [156]

10 Lima beens
8 red beans
2 black-eyed beans

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2 years ago

Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t

krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

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 normally distributed samples are 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.

In this problem, we have that:

$\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

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

By the Central Limit Theorem

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

$Z = \frac{1476 - 1467}{13.2857}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 1458

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

$Z = \frac{1458 - 1467}{13.2857}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

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3 years ago