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2 years ago

Find the opposite, or additive inverse of 3.7

Mathematics

2 answers:

arlik [135]2 years ago

7 0

Answer:

-3.7

Hope that helps!

Step-by-step explanation:

Ymorist [56]2 years ago

7 0

Answer: -3.7!

Expiation!

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25 POINTS HELP NOW<br><br> Twenty is 10% of what number?<br><br> Enter your answer in the box

Alex777 [14]

divide 20 by 10%

10% = 0.10

20 / 0.10 = 200

20 is 10% of 200

3 0

3 years ago

Read 2 more answers

What is the equation of the following graph in vertex form? Courtesy of Texas Instruments A. y = (x − 4)2 − 4 B. y = (x + 4)2 −

Shtirlitz [24]

Answer:

A: y = (x − 4)^2 − 4

Step-by-step explanation:

vertex=(4.-4)

A: y = (x − 4)^2 − 4

y=x^2-8x+16-4

y=x^2-8x+12 (a=1,b=-8,c=12)

the y intercept is (0,12)

vertex ( h, k)

h=-b/2a ⇒ h=-(-8)/2=4

plug the value of h in the equation y=x^2-8x+12

k=4²-8(4)+12

k=16-32+12

k=-4

v(4,-4)

6 0

2 years ago

What is the reciprocal of -b/3

Ludmilka [50]

Answer:

-3/b

Step-by-step explanation:

The reciprocal of -b/3 is -3/b

8 0

3 years ago

The temperature and Williams town is 78 and the temperature in Parkville is 13 cooler what is the temperature in Parkville

kap26 [50]

Answer: 65 degrees

Step-by-step explanation:

Since cooler means lower, then that means subtraction. So, 13 subtracted from 78 is 65.

7 0

2 years ago

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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