The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the systolic reading be the independent (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 146 . Is the predicted value close to 91.0 , which was the actual diastolicreading? Use a significance level of 0.05.Systolic 117 134 150 113 138 113 144 145Diastolic 85 74 86 55 75 80 106 841. What is the regression equation? (Round to two decimal places as needed.)2. What is the best predicted diastolic pressure for a person with a systolic reading of 146 ?3. Is the predicted value close to 91.0 , which was the actual diastolic reading?A. The predicted value is very close to the actual diastolic reading. B. The predicted value is not close to the actual diastolic reading. C. The predicted value is close to the actual diastolic reading. D. The predicted value is exactly the same as the actual diastolic reading.

ŷ = 6.69X - 706.18 ; 270.56; B. The predicted value is not close to the actual diastolic reading.

Step-by-step explanation:

Given the data below :

Systolic reading(x) :

117

134

150

113

138

113

144

145

Diastolic (y) :

85

74

86

55

75

80

106

841

A) Find the regression equation, letting the systolic reading be the independent (x) variable:

Using the online regression calculator :

The regression equation obtained is ;

ŷ = 6.69X - 706.18

Where ;

ŷ = dependent variable

6.69 is the gradient or slope

-706.18 is the intercept, where the line crosses the y - axis.

X - the independent variable (Systolic reading)

B.) Find the best predicted diastolic pressure for a person with a systolic reading of 146 .

ŷ = 6.69X - 706.18

ŷ = 6.69(146) - 706.18

ŷ = 270.56

Is the predicted value close to 91.0

B. The predicted value is not close to the actual diastolic reading.

step-by-step explanation:

i need to answer questions so i dont know but i think its the first one because it goes up but a consent rate

boxplot e is the answer