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2 years ago

Can sum1 do this for me

Mathematics

1 answer:

Ierofanga [76]2 years ago

5 0

Hey the picture is blurry can u take another one plz

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P(x)=x4-52+9x+8 Hallar p(-3)

ozzi

Answer:

P(x)=x⁴+9x-44

Step-by-step explanation:

6 0

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

Figure CDEF is a parallelogram.

NikAS [45]

Answer:

n=10, CF=59, FE=49

6 0

3 years ago

Read 2 more answers

Given f(x) = 4x + 2, find f-1(3).

Alexxandr [17]

Answer:

1/4

Step-by-step explanation:

The inverse function tells you the input value that corresponds to the given output value of the function.

<h3>Application</h3>

You want f^-1(3), which is the value of x such that f(x) = 3.

f(x) = 3 = 4x +2

1 = 4x . . . . . . subtract 2

1/4 = x . . . . . divide by 4

The inverse function of 3 is 1/4:

$\boxed{f^{-1}(3)=\dfrac{1}{4}}$

4 0

2 years ago

Which expression is equivalent to -7(y-2)

soldi70 [24.7K]

Hello!

Distributive property: a(b+c)=ab+ac

$a=-7, b=y, c=2$

$-7y-(-7)*2$

Minus and plus sign.

$-7y+7*2$

Then multiply by the numbers.

$7*2=14$

$\boxed{=-7y+14}$

Hope this helps!

Thanks!

Have a great day!

5 0

3 years ago