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3 years ago

What is the approximate circumference of a circle with a radius of km? Use A. km B. km C. km D. km Question Resources

1 answer:

6 0

In order to calculate the Circumference, You can Use the formula:

C = 2πr
C = πd

They are valid one's, depends, on the given value, either diameter or radius

Hope this helps!

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What is y in 4.58 + y = 2.5

SCORPION-xisa [38]

Answer:

y= 2.08

Step-by-step explanation:

4.58 - 2.5

4.58

- 2.50

-----------

2.08

8 0

3 years ago

Read 2 more answers

Kloe has 24 sunflowers, 16 roses, and 48 white carnations. She wants to make flower arrangements with the same number of each fl

Veseljchak [2.6K]

Kloe can make 4 flower arrangement

The number of flowers in each flower arrangement is 6 sun flowers, 4 roses and 12 white carnations.

The number of flower arrangements can be calculated as follows;

The first step is to calculate the HCF between the three numbers

24= 6 × 4

16= 4 × 4

48= 12 × 4

The HCF is 4

Hence the number of flower arrangement is 4 and the number of flowers in each flower arrangement is 6 sun flowers, 4 roses and 12 white carnations

Read more on flowers here

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1 year ago

What is the interquartile range of the following data set? 78,90,456,676,111,381,21

Fiesta28 [93]

Think you gotta add them all up them divide by how many numbers their are

7 0

4 years ago

Leslie invested in an account that paid 2% annual interest compounded continuously. After 40

daser333 [38]

40 x 3^ djwk17393959
Ow1939499292919

3 0

3 years ago

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (

alisha [4.7K]

Answer:

Point estimation of the mean = 17.6

Error = 4.6

95% CI: 13.0 ≤ μ ≤ 22.2

The distribution can be approximated to a normal because the sample size n=50 is bigger than 30.

Step-by-step explanation:

We have this information:

- Sample mean:

$\bar{X}=17.598$

- Sample standard deviation

$s=16.01712719$

- Sample size

$n=50$

For now on, we round to one decimal place.

The best estimation for the population mean is the sample mean

$\mu=\bar{X}=17.6$

The margin of error is equal to the t-value multiplied by the sample standard deviation and divided by the sample size.

The t value depends on the degrees of freedom and the width of the confidence interval. In this case it is a 95% CI and the degrees of freedom are 49. For this conditions, the t-value is t=2.01.

Then, the margin of error is

$E=t_{49}*\sigma/\sqrt{n}=2.01*16.0/\sqrt{50}=4.553$

Then, the confidence interval can be constructed as:

$\bar{X}-t\sigma/\sqrt{n}\leq\mu\leq\bar{X}-t\sigma/\sqrt{n}\\\\17.6-4.6\leq \mu \leq 17.6+4.6\\\\ 13.0\leq \mu \leq 22.2$

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3 years ago