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2 years ago

What is the area of a sector with a central angle of 5pi over 6 radians and a radius of 5.6ft

Mathematics

1 answer:

kogti [31]2 years ago

6 0

Answer:

41.0501 ft²

Step-by-step explanation:

Area of a Sector (Radians): A = 1/2r²∅

We are given r and ∅, so simply plug it into the formula:

A = 1/2(5.6)²(5π/6)

A = 1/2(31.36)(5π/6)

A = 15.68(5π/6)

A = 41.0501 ft²

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an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

--------------------------------

We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score 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, which is the percentile of X.

In this problem:

Mean of 8.8 inches, thus $\mu = 8.8$ .
Standard deviation of 2.8 inches, thus $\sigma = 2.8$ .

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

The diameter is 1.5IQR above the 75th percentile, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

A similar problem is given at brainly.com/question/15683591

3 0

1 year ago

Find the sum of the series given in sigma notation.

Kryger [21]

Answer:

Option b is correct 175

Step-by-step explanation:

n = 7

k = 6

3k -2 ------1

put k = 6 in above eq. for finding first term

a1 = 3(6) - 2 = 18 - 2 = 16

put k = 7 in above eq. for finding first term

a2 = 3(7) - 2 = 21 - 2 = 19

a3 = 3 (8) - 2 = 24 - 2 = 22

16, 19 , 22, ... //Arithmetic series formation

a1 = 16 , a2 = 19

d = a2 - a1 = 19 - 16 = 3 //Difference of first two terms

Using sum forumula for arithmetic series

sum = $\frac{n}{2}(2a1 + (n-1)d)$

= $\frac{7}{2}(2a1 + (7-1)d)$

= $\frac{7}{2}(2(16) + (6)3)$

= $\frac{7}{2}(32 + 18)$

= $\frac{7}{2}(50)$

= 7 * 25

= 175

4 0

2 years ago

Suppose yoko borrows $3500 at an interest rate of 4% compounded each year Find the amount owed at the end of 1 year

olchik [2.2K]

Amount owed at the end of 1 year is 3640

<h3>Solution:</h3>

Given that yoko borrows $3500.

Rate of interest charged is 4% compounded each year

Need to determine amount owed at the end of 1 year.

In our case :

Borrowed Amount that is principal P = $3500

Rate of interest r = 4%

Duration = 1 year and as it is compounded yearly, number of times interest calculated in 1 year n = 1

Formula for Amount of compounded yearly is as follows:

$A=p\left(1+\frac{r}{100}\right)^{n}$

Where "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula we get

$\mathrm{A}=3500\left(1+\frac{4}{100}\right)^{1}$

$\begin{array}{l}{A=\frac{3500 \times 104}{100}} \\\\ {A=35 \times 104} \\\\ {A=3640}\end{array}$

Hence amount owed at the end of 1 year is 3640

5 0

2 years ago

What is the formula to find the SA of a pyramid

shusha [124]

Answer:

SA= 12pl+B

Step-by-step explanation:

8 0

3 years ago

Is y=3/4x a direct variation? if so, find the constant of variation

xxMikexx [17]

Yes because without a y-intercept it goes through 0. and can you explain the next part?

8 0

3 years ago