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3 years ago

−5x−9y=3 plus 5x−9y=-2

Mathematics

1 answer:

Murrr4er [49]3 years ago

5 0

Answer:

-18y=1

Step-by-step explanation:

Sorry if this is wrong, I was just guessing...

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Question 9 (1 point)

Romashka [77]

Answer:

Step-by-step explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(82.5-1.5\times 76.56,159.06+1.5\times 76.56)$

(-32.34,273.9)

So, There is no outlier

Maximum = 207

League B in ascending order :

24.93,136,145,157,163.25,178,256

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

$Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5$

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(140.5-1.5\times 30.125,170.625+1.5\times 30.125)$

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256

5 0

3 years ago

15(3x - 4y)<br> Use Distributive property

MariettaO [177]

45x -4y is the answer

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3 years ago

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ASAP HELP ASAP THANK YOU

aliya0001 [1]

15.6/-3 = - 5.2 is the answer

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3 years ago

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The length of a rectangle is 12 in and the perimeter is 56 what us the width of the rectangle

Aleks [24]

Hi there!

We can find the perimeter of a rectangle by using the following formula:

perimeter = 2 × width + 2 × length

In the question, we are given the following data: the length of the rectangle is 12 in and the perimeter is 56. Let's substitute this into our formula!

56 = 2 × width + 2 × 12
Multiply first.

56 = 2 × width + 24
Now subtract 24 from both sides.

32 = 2 × width
And finally, to find the width of the rectangle, divide both sides of the equation by 2.

16 = width
(we can eventually switch sides in the equation).

width = 16
~ Hope this helps you!

3 0

3 years ago

What is the LCM<br> OF 4, 8 and 16

sleet_krkn [62]

I think the answer is 2

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3 years ago

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−5x−9y=3 plus 5x−9y=-2 ​

−5x−9y=3 plus 5x−9y=-2