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EastWind [94]
2 years ago
15

Please help me with my delta math work

Mathematics
1 answer:
Luden [163]2 years ago
7 0

Answer:

use photo math it's free and easy

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Why is circle 1 similar to circle 2?
Artemon [7]
Circle 2 is a dial at ion of circle 1
3 0
2 years ago
What value of b will cause the system to have an infinite number of solutions?
Leya [2.2K]

Answer:

-6

Step-by-step explanation:

V = 6x + b

 1/2 V -3 x = -3

V - 6x = -6

V - 6x  =  b

3 0
2 years ago
There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s
harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

     Original Mean = $70,000

       \frac{\sum X}{n} = $70,000  ⇒ \sum X = 70,000 * 10 = $700,000

   New \sum X after change = $700,000 - $100,000 + $1,000,000 = $1600000

  Therefore, New mean = \frac{1600000}{10} = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = 20000^{2} i.e.  20000^{2} = \frac{\sum x^{2} - n*xbar }{n -1}

    Original \sum x^{2} = (20000^{2} * (10-1)) + (10 * 70,000) = $3,600,700,000

    New \sum x^{2} = $3,600,700,000 - 100,000^{2} + 1,000,000^{2} = 9.936007 * 10^{11}  

    New Variance = \frac{new\sum x^{2} - n*new xbar }{n -1} = \frac{9.936007 *10^{11}  - 10*160000 }{10 -1} = 1.103999 * 10^{11}    Therefore, standard deviation after change = \sqrt{1.103999 * 10^{11} } = $332264.804 .

7 0
3 years ago
PLEASE HELP/ANSWER! From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 198
Elena-2011 [213]

The weight in 1980 is \frac{4k}{5} kilograms

<em><u>Solution:</u></em>

From 1980 to 1990, Lior’s weight increased by 25%

His weight is "k" kilograms in 1990

<em><u>To find: weight in 1980</u></em>

This is a percentage increase problem

Let "x" be the weight in kilograms in 1980

<em><u>The percentage increase is given by formula:</u></em>

\text{Percentage increase } = \frac{\text{Final value - initial value}}{\text{initial value}} \times 100

Here,

Initial value in 1980 = x

Final value in 1990 = k

Percentage increase = 25 %

<em><u>Substituting the values in formula,</u></em>

25 = \frac{k-x}{x} \times 100\\\\25x = 100(k-x)\\\\x = 4(k-x)\\\\x = 4k - 4x\\\\5x = 4k\\\\x = \frac{4k}{5}

Thus the weight in kilograms in 1980 is \frac{4k}{5}

7 0
3 years ago
Can you solve some of these? Or all, whichever one you might know. Also explain please.
Rina8888 [55]
Here's number 5. If you need more explanation just let me know.

8 0
3 years ago
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