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2 years ago

Can someone help me with this?

Mathematics

1 answer:

mr Goodwill [35]2 years ago

7 0

If you look closely you'll see that x can take on any real value here. On the other hand, y can equal -4 (maximum) or any real number smaller than -4.
Translate this information into interval notation.

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Need help kinda quick

Fed [463]

Answer:

A;-6

B;-8

C;-2

D;-14

E;-1

F;2

G;5

H;0

4 0

2 years ago

Sam brought brushes for $8,a palette for $5,and oil paints for $15.He paid 29.82 in all.What sale-tax rate did Sam pay?

sertanlavr [38]

8+5+15=28
29.82-28= 1.82 or 182%

5 0

3 years ago

2. In your own words, explain HOW to find the standard deviation of this data set. (20 points) 3. What does the standard deviati

Ket [755]

Answer:

$\sigma = 11.28$ --- Standard deviation

Step-by-step explanation:

Given

See attachment for graph

Solving (a): Explain how the standard deviation is calculated.

<u>Start by calculating the mean</u>

To do this, we divide the sum of the products of grade and number of students by the total number of students;

i.e.

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{50 * 1 + 57 * 2 + 60 * 4 + 65 * 3 +72 *3 + 75 * 12 + 77 * 10 + 81 * 6 + 83 * 6 + 88 * 9 + 90 * 12 + 92 * 12 +95 * 2 + 99 * 4 + 100 * 5}{1 + 2 + 4 + 3 +3 + 12 + 10 + 6 + 6 + 9 + 12 + 12 +2 + 4 + 5}$

$\bar x = \frac{7531}{91}$

$\bar x = 82.76$

Next, calculate the variance using the following formula:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f}$

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset

So, we have:

$\sigma^2 = \frac{1 * (50 - 82.76)^2 +...................+ 5 * (100 - 82.76)^2}{91}$

$\sigma^2 = \frac{11580.6816}{91}$

$\sigma^2 = 127.26$

Lastly, take the square root of the variance to get the standard deviation

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{127.26}$

$\sigma = 11.28$ --- approximated

<em>Hence, the standard deviation is approximately 11.28</em>

Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.