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kogti [31]
3 years ago
14

Mrs. Morales wrote a test with 16 questions covering spelling and vocabulary. Spelling questions (x) are

Mathematics
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

<h3>5x+10y=100</h3>

Step-by-step explanation:

X+Y = 15

<h3>Sum of two types of question is 15</h3>

x+y-x = -x+15

<h3>y= -x+15</h3>

5(x)+10(y)=100

5x+10y=100

<h3>Total no. if points is 100</h3>

5x+10y-5x=-5x+100

10y= -5x+100

y=5/10x + 100/10

y= -1/2x+10

<h3>Slope intercept form</h3>

(10,5)

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A sample of blood pressure measurements is taken for a group of​ adults, and those values​ (mm Hg) are listed below. The values
Slav-nsk [51]

Answer:

Systolic on right

\hat{CV} =\frac{18.68}{127.5}=0.147

Systolic on left

\hat{CV} =\frac{12.65}{74.2}=0.170

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

Step-by-step explanation:

Assuming the following data:

Systolic (#'s on right) Diastolic (#'s on left)

117; 80

126; 77

158; 76

96; 51

157; 90

122; 89

116; 60

134; 64

127; 72

122; 83

The coefficient of variation is defined as " a statistical measure of the dispersion of data points in a data series around the mean" and is defined as:

CV= \frac{\sigma}{\mu}

And the best estimator is \hat {CV} =\frac{s}{\bar x}

Systolic on right

We can calculate the mean and deviation with the following formulas:

[te]\bar x = \frac{\sum_{i=1}^n X_i}{n}[/tex]

s= \frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}

For this case we have the following values:

\bar x = 127.5, s= 18.68

So then the coeffcient of variation is given by:

\hat{CV} =\frac{18.68}{127.5}=0.147

Systolic on left

For this case we have the following values:

\bar x = 74.2 s= 12.65

So then the coeffcient of variation is given by:

\hat{CV} =\frac{12.65}{74.2}=0.170

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

4 0
3 years ago
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