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3 years ago

(-11x+31y) - 2(-x + 5y)

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

Equivalent expression of (-11x+31y) - 2(-x + 5y) is -9x +21y Hence option C is correct

Solution:

Given expression is (-11x+31y) - 2(-x + 5y)

Need to find equivalent expression from four given option.

Let’s first simplify the given expression

(-11x+31y) - 2(-x + 5y)

On opening the brackets we get

-11x + 31y + 2x -10y

Now bringing similar terms together and performing appropriate operation, we get

-11x + 2x + 31y + -10y

Taking common terms out we get,

=> (-11 + 2) x + (31 -10) y

= -9x +21y

Therefore Equivalent expression of (-11x+31y) - 2(-x + 5y) is -9x +21y.

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12 + 54=6(_+_) please help me 14 points but please just help me

sp2606 [1]

Are there any other details as to what the numbers are? otherwise there are six possibilities for this
12+54=66
66=6(_+_)
11=(_+_)

1) 0, 11
2) 1, 10
3) 2, 9
4) 3, 8
5) 4, 7
6) 5, 6

7 0

3 years ago

Read 2 more answers

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

Tom, Lisa, and Jane are competing in a cooking competition. They all used different amounts of sugar from a can containing 4 pou

avanturin [10]

Convert all to decimals
percent means parts out of 100
13%=13/100=0.13

'of' means mutiply

Tom=13% of 4
tom=0.13 times 4
tom=0.52 lb

Lisa=0.6 lb

Jane=0.12 of 4
jane=0.12 times 2
jane=0.48

Tom=0.52
Lisa=0.60
Jane=0.48

Lisa used the most

8 0

3 years ago

Please simplify this expression ! Show your work and please answer this !!

tangare [24]

Answer:

18q^2 - 52q + 32

Step-by-step explanation:

We can use the Distributive Property to solve this.

(2q-4)(9q-8)

2q(9q) + 2q(-8) + -4(9q) + -4(-8)

18q^2 - 16q - 36q + 32

18q^2 - 52q + 32

18q^2 - 52q + 32

6 0

3 years ago

Help I'll give extra cred if u do

Schach [20]

I think it’d be 2 and 3 if you’re rounding 2.984

7 0

2 years ago