Answer:
Hence, the data provides convincing evidence that a linear relationship exists between hours of sleep observed and academic performance as measured by GPA.
Step-by-step explanation:
Given the data:
Sleep (hrs) 9 8.5 9 7 7.56 7 5.5 6 8.5 6.5 8
GPA 3.8 3.3 3.5 3.6 3.4 3.3 3.2 3.2 3.2 3.4 3.6 3.1 3.4 3.7
The scatter plot shows a positive linear trend. With the correlation Coefficient depicting a R value of 0.56. The residual plot also depicts a a randomly scattered values of the residual values. Similarly, a plot of the normal values of residuals
Answer: 1.44 x 10-7 m5
Step-by-step explanation:
(8 cm) x (6 cm) x (3 cm) x (5 cm) x (2 cm) =
1.44 × 10-7 m5
I’m really sorry if it’s not correct 0_O
Answer:
h = f/7g
Step-by-step explanation:
You have to rearrange the equation so that h is alone on one side.
f = 7gh
f/7g = (7gh)/(7g)
f/7g = h
h = f/7g
So you first multiply the fractions: 2/5 times -1/5 = -2/25; then the a's: a^2 times a equals a^3; then multiply the x's: x times x^2 equals x^3.
The answer would be -2/25a^3x^3
Answer:
(8.213 ; 8.247)
Step-by-step explanation:
Given the data :
No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Dia. 8.23 8.16 8.23 8.25 8.26 8.23 8.20 8.26 8.19 8.23 8.20 8.28 8.24 8.25 8.24
Sanple size, n = 15
Sample mean, xbar = Σx / n = 123.45 / 15 = 8.23
The sample standard deviation, s = √(x -xbar)²/n-1
Using calculator :
Sample standard deviation, s = 0.03116
s = 0.031 (3 decimal places)
The 95% confidence interval :
C.I = xbar ± (Tcritical * s/√n)
Tcritical at 95%, df = 15 - 1 = 14
Tcritical = 2.145
C.I = 8.23 ± (2.145 * 0.031/√15)
C.I = 8.23 ± 0.0171689
C.I = (8.213 ; 8.247)
