What is the area of the figure?

Solve each equation for the specified variable y+v=w solve for v

nasty-shy [4]

Answer: v=w-y

Step-by-step explanation:

y+v=w

-y -y

v=w-y

4 0

3 years ago

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Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re

mixer [17]

Answer: range= 26, variance= 80 and standard deviation= 8.94

Step-by-step explanation:

Range = highest - lowest

Range = 146 - 120

Range= 26

Let m be mean

M=mean=sum/n

Mean=(120+134+146+127+138+133) / 6

M=798/6

M=133

The standard deviation sample formula:

S.D = sqrt( Summation of |x-m|^2 / n-1)

Let start finding:

|x-m|^2

For 1st: |120-133|^2=169

For 2nd: |134-133|^2=1

For 3rd: |146-133|^2=169

For 4th: |127-133|^2=36

For 5th: |138-133|^2=25

For 6th: |133-133|^2=0

Summation of |x-m|^2 = 400

The standard deviation formula is :

S.D = sqrt( Summation of |x-m|^2 / n-1)

S.D= sqrt(400 / 5)

S.D=sqrt(80)

S.D= 8.94

Variance = (Summation of |x-m|^2 / n-1)

Variance= 400/5

Variance= 80

3 0

3 years ago

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Area of a regular octagon with radius 8

bixtya [17]

Step-by-step explanation:

Area of the octagon = 2(1+√2)8=309.02 units

3 0

3 years ago

Solve the equation using the quadratic formula : x^2-10x+25=18

Sever21 [200]

$x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}$

$Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}$

8 0

3 years ago

the first group of students consists of 10 and their average age was 13 years old the next group consisted of 30 students and th

Kazeer [188]

Answer: The average was 9 years old find the average age of both groups is 10 years old.

Step-by-step explanation:

Formula foe average : $\text{Average} =\dfrac{\text{Sum of all observations}}{\text{Number of observations}}$

Given : The first group of students consists of 10 and their average age was 13 years old.

i.e. $13=\dfrac{\text{Sum of ages of students of first group}}{10}\\\\\Rightarrow\ \text{Sum of ages of students of first group}=13\times10=130$ (1)

The next group consisted of 30 students and their average was 9 years old.

i.e. $9=\dfrac{\text{Sum of ages of students of next group}}{30}\\\\\Rightarrow\ \text{Sum of ages of students of nextgroup}=9\times30=270$ (2)

Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400

Combined students of both groups (first and next group )= 10+30=40

Now , the average of both groups = $\dfrac{\text{Sum of combined students}}{\text{combined students}}$

$=\dfrac{400}{40}=10$

Hence, the average was 9 years old find the average age of both groups is 10 years old.

4 0

3 years ago