Answer:
a. H0:μ1≥μ2
Ha:μ1<μ2
b. t=-3.076
c. Rejection region=[tcalculated<−1.717]
Reject H0
Step-by-step explanation:
a)
As the score for group 1 is lower than group 2,
Null hypothesis: H0:μ1≥μ2
Alternative hypothesis: H1:μ1<μ2
b) t test statistic for equal variances
t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}
t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}
t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]
t=-3.076
c. α=0.05, df=22
t(0.05,22)=-1.717
The rejection region is t calculated<t critical value
t<-1.717
We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.
Answer:
Volume = 16 unit^3
Step-by-step explanation:
Given:
- Solid lies between planes x = 0 and x = 4.
- The diagonals rum from curves y = sqrt(x) to y = -sqrt(x)
Find:
Determine the Volume bounded.
Solution:
- First we will find the projected area of the solid on the x = 0 plane.
A(x) = 0.5*(diagonal)^2
- Since the diagonal run from y = sqrt(x) to y = -sqrt(x). We have,
A(x) = 0.5*(sqrt(x) + sqrt(x) )^2
A(x) = 0.5*(4x) = 2x
- Using the Area we will integrate int the direction of x from 0 to 4 too get the volume of the solid:
V = integral(A(x)).dx
V = integral(2*x).dx
V = x^2
- Evaluate limits 0 < x < 4:
V= 16 - 0 = 16 unit^3
The grade of a road is the slope of a road. In the U.S., grade is often expressed as a percent by finding the product 100(slope). Approximate the grade of a road that has a rise of 950 ft over 3 mi is :
A. 3%
Its depends on how many times you score during the game
in second game the number of points increases as a geometric sequence
with common ratio 2
so for example if you score ten times in first game you get 2000 points
if you score 10 times in Game 2 you score 2 * (2^10) - 1 = 2046 points
so playing 10 or more games is best with Game 2. Any less plays favours gane 1.
