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kolbaska11 [484]
3 years ago
14

A department of transportation's study on driving speed and miles per gallon for midsize automobiles resulted in the following d

ata. Speed (Miles per Hour) 30 50 40 55 30 25 60 25 50 55 Miles per Gallon 29 26 26 22 29 31 21 35 25 26 Compute the sample correlation coefficient. (Round your answer to two decimal places.) Interpret the sample correlation coefficient. There is a strong negative relationship between the driving speed and miles per gallon. For driving speed between 25 and 60 miles per hour, higher speeds are associated with lower miles per gallon.
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Answer:

R = -0.90 (2 decimal places)

Step-by-step explanation:

Given the data :

Speed (Miles per Hour) :

30

50

40

55

30

25

60

25

50

55

Miles per Gallon :

29

26

26

22

29

31

21

35

25

26

The correlation Coefficient, R shows the degree or level of relationship between the dependent and independent variable. A positive R value denotes a positive correlation while a negative R value denotes a negative relationship. R values should range between - 1 to +1 with values close to 1 and - 1 depicting strength while value of zero means no relationship exist.

Hence, R value of -0.90 depicts a strong negative relationship. Hence, higher speed are related to lower miles per gallon and vice versa.

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What is the following product? fourth root of 7 times the fourth root of 7
Cerrena [4.2K]

Answer:

square root of seven

Step-by-step explanation:

multiply the numbers inside the radicals to get fourth root of 49. simplify the radical and get square root of 7

3 0
3 years ago
Consider these four data sets. Set A: {32, 12, 24, 46, 18, 22, 14} Set B: {4, 12, 11, 14, 11, 5, 12, 13, 18, 14} Set C: {5, 4, 9
drek231 [11]
SET A

The elements of this set are,

32,12,24,46,18,22,24


We arrange this data set in ascending order to get,

12,18,22,24,24,32,46



The median for this data set is

median = 24


The mean for this data set can be calculated using the formula,

\bar X =  \frac{ \sum x}{n}

This implies that,

\bar X =  \frac{ 12 + 18 + 22 + 24 + 24 + 32 + 46}{7}

\bar X =  \frac{154}{7}  = 22
This data set is negatively skewed because the mean is less than the median.



SET B

This set contains the elements,

4, 12, 11, 14, 11, 5, 12, 13, 18, 14



We arrange the elements of this set in ascending order to obtain,

4,5,11,11,12,12,13,14,14,18

The median for this data set is,

median =  \frac{12 + 12}{2}  = 12

The mean is

\bar X =  \frac{4 + 5 + 11 + 11 + 12 + 12 + 13 + 14 + 14 + 18}{10}

.
\bar X  =  \frac{114}{10}  = 11.4

This data set is negatively skewed because the mean I'd less than the median.

SET C

The set C has elements,
5, 4, 9, 12, 14, 26, 22, 18

We arrange this set in ascending order to get,

4,5,9,12,14,18,22,26

The median of this data set is

median =  \frac{12 + 14}{2}  = 13


The mean of this data set is
\bar X =  \frac{4 + 5 + 9 + 12 + 14 + 18 + 22 + 26}{8}


\bar X =  \frac{110}{8}  = 13.75


This is a positively skewed distribution because the mean is greater than the median


SET D


The elements of this set are

1, 1, 1, 2, 2, 3, 3, 4, 5, 6

The median of this data set is

median =  \frac{2 + 3}{2}  = 2.5


The mean is
\bar X= \frac{1 + 1 + 1 + 2 + 2 + 3 + 3 +  4+ 5 + 6}{10}


\bar X= \frac{28}{10}  = 2.8

This data set is positively skewed because the mean is greater than the median.

4 0
3 years ago
Read 2 more answers
There are 15 boys in school base ball team , they represent 5% of grace Christian school , In all how many are there in school
serious [3.7K]

Answer:

300

Step-by-step explanation:

5\% \: of  \: x = 15 \\ x = 1500 \div 5 = 300

6 0
2 years ago
PLS HELP !!!!
Masteriza [31]
(300 students) because each classroom has 25 students then you times that by 12 and you get 300.
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3 years ago
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Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0
nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for  x  are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit  (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term  =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let  a0  and  a1  be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since  a0=4a2 , I’ll choose  a1  and  a2  as the two arbitrary constants. We can still express all other constants in terms of  a1  and/or  a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all  n≥0 , except for  n=1 , since  a1  is an arbitrary constant independent of  a0  (and therefore independent of  a2 ).

Plugging back into original power series for  y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant  a2  is  =4+  a power series (starting at  n=2 ). However, if we had the appropriate  x -term, we would have a power series starting at  n=0 . Since the other independent solution is simply  y1=x,  then we can let  a1=c1−3c2,   a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

brainly.com/question/11443401

#SPJ4

6 0
1 year ago
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