<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bm%7D%7B%20-%207%7D%20%20%3C%202" id="TexFormula1" title=" \frac{m}{

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Ratling [72]

3 years ago

alt=" \frac{m}{ - 7} < 2" align="absmiddle" class="latex-formula">
what do I do hfhr

Mathematics

1 answer:

aksik [14]3 years ago

8 0

Answer:

$m = -14$

Step-by-step explanation:

Solved for $m$ ! If you need the inequality to $m$ I'd love to help!

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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

What does a trapezium look like

sasho [114]

This is a trapezium with congruent unparallel lines.

5 0

3 years ago

What’s the average of those numbers stated below.

Sedaia [141]

Answer:

Step-by-step explanation:

I counted any number that had the tens place <u>80</u>

7 0

2 years ago

Read 2 more answers

PLEASE HELP SHOW YOUR WORKING OUT Branliest

Ad libitum [116K]

Answer:

The equation of this line is therefore y = 2x + 3.

Step-by-step explanation:

this line passes thru the points (0, 3) and (3, 9). As we move from (0, 3) to (3, 9), x increases by 3 and y increases by 6. Thus, the slope of this line is

m = rise / run = 6/3, or m = 2. Inserting the known info (m = 2, x = 0, y = 3) into y = mx + b, we get: 3 = 2(0) + b, so we see that b = 3.

The equation of this line is therefore y = 2x + 3.

5 0

3 years ago

The base of a parallelogram is28.4 centimeters the height is one fourth of the base. What the area of the parallelogram?

brilliants [131]

The base of a parallelogram is 28.4 cm.
The height is one fourth of the base.
28.4 ÷ 4 = 7.1, so the height is 7.1 cm.
What the area of the parallelogram?
The area of any parallelogram is the base times the height.
28.4 cm × 7.1 cm = 201.64 cm²

4 0

3 years ago

Read 2 more answers