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3 years ago

PLEASE HELP ME I BEGGGGGGG

Mathematics

2 answers:

Hoochie [10]3 years ago

6 0

Answer:

Step-by-step explanation:

marissa [1.9K]3 years ago

4 0

Answer:

Step-by-step explanation:

I believe it is A because according to the theorem it states the other pair of opposite sides are parallel.

"Here's the game plan. Assume that ¯BC ‌ ‌ ¯AD and ¯BC ~= ¯AD. By definition, a parallelogram is a quadrilateral with both pairs of opposite sides parallel. You already know that one pair of opposite sides is parallel. You need to show that the other pair of opposite sides is parallel. In other words, you need to show that ¯AB ‌ ‌ ¯CD."

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What is the value of 166

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4 years ago

Need help. I can’t fail this!! 7

vodka [1.7K]

Answer:

Third one? I'm not sure though.

Step-by-step explanation:

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3 years ago

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

For which value of k does the equation |14 + x| -5 = k have exactly one solution

Blizzard [7]

Answer:

k = |14 + x| -5

Step-by-step explanation:

4 0

3 years ago

If $a$, $b$, $c$, and $d$ are replaced by four distinct digits from $1$ to $9$, inclusive, then what's the largest possible valu

Alexxx [7]

Answer:

Step-by-step explanation:

for the value of difference to be largest, the minuend should be maximum(most possibly) and the subtrahend should be minimum

[in A-B=X, A is minuend and B is subtrahend ]

so, $a.b should be maximum. as there is a condition that 4 digits should be distinct, the product will be maximum if we choose 2 maximum valued numbers from the given numbers. so, one of them should be 9 and the other should be 8.

therefore, $a.b=9*8=72

as mentioned above, c.d$ should be minimum. this will be possible only when we choose 2 minimum valued numbers from the given numbers. so, one of them should be 1 and the other should be 2.

therefore, c.d$ = 1*2 = 2

hence, the difference = 72-2 = 70

thus, the largest possible value of the difference $a.b - c.d$ = 70

7 0

3 years ago