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2 years ago

How do you find the mean, median and mode?

2 answers:

6 0

The median is the number in the middle of the set of data
ex. 1, 2, 3 - 2 is the median
ex. in some cases, the median is the sum of the two numbers in the middle of set of data over 2
The mode is the number that is repeated in the set of data
ex. 1, 1, 2, 2 - 1 and 2 would be the mode
The mean is the sum of all numbers divided by how many number there are
ex. 1, 2, 3, 4 - 1+2+3+4/2
you didn't ask but I'm sure you'll come to it
so the range is the difference of the smallest value and the largest value in a set of data
ex. 1, 2, 3 - range would be 2

algol132 years ago

3 0

If there is an odd number of data values then the median will be the value in the middle. If there is an even number of data values the median is the mean of the two data values in the middle. For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5. For the data set 1, 1, 2, 6, 6, 9 the median<span> is 4.
</span>The mean<span> is the average of the numbers: a calculated "central" value of a set of numbers. To calculate: Just add up all the numbers, then divide by how many numbers there are.</span><span>

</span>

You might be interested in

Please help me out :P

Veseljchak [2.6K]

cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = $\sqrt{7^{2} + 4^{2} }$ = $\sqrt{49+16}$ = $\sqrt{65}$

sin θ = opposite/ hyp = $\frac{7}{\sqrt{65} }$

Rationalize, $\frac{7}{\sqrt{65} }$ x $\frac{\sqrt{65} }{\sqrt{65} }$ = $\frac{7\sqrt{65} }{65}$

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - $\frac{7\sqrt{65} }{65}$

cos θ = adj/hyp = $\frac{4}{\sqrt{65} }$

By rationalizing and knowing that cos θ is negative, cos θ = - $\frac{-4\sqrt{65} }{65}$

cot θ = 1/tan θ = 1/7/4 = 4/7

sec θ = 1/cos θ = 1/ $\frac{4}{\sqrt{65} }$ = - $\frac{-\sqrt{65} }{4}$

cosec θ = 1/sin θ = 1/ $\frac{\sqrt{65} }{7}$ = $\frac{-\sqrt{65} }{7}$

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0

1 year ago

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. Thi

Dovator [93]

Complete question:

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.

A) Find the mean and standard deviation for the number of girls in groups of 36 births.

B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.

C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?

Answer:

a) mean = 18

Standard deviation =3

b) low range = 12

High range = 24

c) The result of 33 girls is significantly high. Yes, the method is effective.

Step-by-step explanation:

Given:

p = 0.5

n = 36

a) The mean is the product of n and p

Mean u = np

u = 36 * 0.5 = 18

The standard deviation is the square root of the product of n and p&q.

S.d ó = $\sqrt{npq}$

$= \sqrt{np(1-p)}$

$= \sqrt{36(0.5)(1-0.5)} = \sqrt{9} = 3$

b) To find the range rule of thumb:

• For low range

Low range = u - 2ó

= 18 - (2 * 3)

= 12

• High range

= u + 2ó

= 18 + (2*3)

= 24

c) The result is significantly high, because 33 is greater than 24 girls.

A result of 33 girls would prove the method as effective.

8 0

2 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$