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ASHA 777 [7]
3 years ago
13

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

adings​ (in mmHg) are listed below. Find the​ range, variance, and standard deviation for the given sample data. If the​ subject's blood pressure remains constant and the medical students correctly apply the same measurement​ technique, what should be the value of the standard​ deviation?120 134 146 127 138 133
Mathematics
2 answers:
vampirchik [111]3 years ago
7 0

Answer:

Part A.

Range= 26

Variance= 66.67

Standard deviation= 8.165

Part B.

If the blood pressure remains constant and all the six medical student gets the same result from the blood pressure, The standard deviation will be; The value which is the same minus the average value.

Step-by-step explanation:

The values of their result are 120, 134, 146, 127, 138, and 133

FIND RANGE:

The range is the highest value minus the lowest value

146 is the highest value

120 is the lowest value

146-120= 26

FIND VARIANCE :

Variance is the mean value of the sum of the square of the difference between each value and the mean.

Variance= ∑[( x- X)]/n

Step 1: find the mean

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

= 133

Step2: square the difference between each value and the mean.

(120-133)^2=169

(134-133)^2=1

(146-133)^2=169

(127-133)^2=36

(138-133)^2=25

(133-133)^2=0

Step 3: find the average of the values in step 2.

(169+1+169+36+25+0)/6

= 400/6 = 66.67

FIND THE STANDARD DEVIATION:

The standard deviation is the square root of the variance.

√66.67= 8.165.

IF THE VALUES OF THEIR RESULT ARE THE SAME, STANDARD DEVIATION WILL BE

x - X

Where;

x is the value of the number which is the same for the six student

X is the mean value of the number which is the same for the six student

mixer [17]3 years ago
3 0

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

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100 POINTS AND BRAINLIYEST!!!
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See explanations below

Step-by-step explanation:

Chanel

(i) 90° counterclockwise and then dilation by a factor of 1/2

Trapezoid PQRS is given as :

P(6,-4) -------P'(4,6)

Q(2,-2)------Q'(2,2)

R(2,-6)-------R'(6,2)

S(6,-6)-------S'(6,6)

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B. The new coordinates are : P'(4,6), Q'(2,2),R'(6,2) and S'(6,6)

The  rule for rotation 90° counterclockwise about the origin is (x,y)⇒(-y,x)

The rule for dilation with scale factor of 1/2 is (x,y)------(1/2x,1/2y)

P'(4,6) -----P''(2,3)

Q'(2,2)-----Q''(1,1)

R'(6,2)-----R''(3,1)

S'( 6,6)-----S''(3,3)

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Preston

E. (i)Reflection over x-axis where the rule is : (x,y)----(x,-y)

(ii)Dilation by scale factor of 1/2 the rule is : (x,y)---(1/2x, 1/2y)

Applying reflection over x-axis rule;

P(6,-4) -----P'(6,4)

Q(2,-2)-----Q'(2,2)

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S(6,-6)------S'(6,6)

The coordinates have changed in that a negative sign is added to the y-axis coordinate.

F. P'(6,4), Q'(2,2), R'(2,6) and S'(6,6)

G. Applying the dilation by scale factor of 1/2

P'(6,4)-------P''(3,2)

Q'(2,2)------Q''(1,1)

R'(2,6)------R''(1,3)

S'(6,6)------S''(3,3)

The coordinates of P'Q'R'S are divide by 2 to give coordinates of P''Q''R''S''

Comparing the answers to the final figure given

K(3,2)

L(1,1)

M(1,3)

N(3,3)

Preston is correct because his sequence proves trapezoid PQRS is similar to trapezoid KLMN.

Learn More

Reflection and dilation :brainly.com/question/13408902

Keywords :trapezoid, maps, rotate, counterclockwise, dilate,reflect,scale factor,shape, coordinate

#LearnwithBrainly

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